Homological Algebra I: Complexes and Resolutions introduces the basic language and methods used to study algebraic objects through chains of modules and maps between them. The course focuses on chain complexes, exactness, and the idea that many structural questions can be reformulated in terms of how kernels, images, and cokernels fit together inside diagrams. From there, it develops the standard diagram-chasing tools that make homological arguments effective, and it shows how homotopy and quasi-isomorphisms capture when two complexes should be regarded as equivalent from a homological point of view. A central theme of the course is extracting information from exact sequences and using it to compare algebraic structures. The early chapters build the formal foundation: complexes, exactness, the snake lemma, the five lemma, and related results. The middle chapters then move to short exact sequences of complexes, chain homotopy, and mapping cones, which together explain how maps of complexes encode homological data. The final chapters turn to projective and injective modules, then to projective and injective resolutions, which provide systematic ways to compute and organize derived information. The chapters are arranged to move from local manipulations of diagrams to global constructions of resolutions. First, students learn to work confidently with exact sequences and commutative diagrams; then they learn to compare complexes up to homotopy and quasi-isomorphism; finally, they apply these ideas to projective and injective objects, where resolutions become the main tool for homological computation. By the end of the course, the basic formalism of homological algebra is in place, preparing for derived functors and more advanced homological methods. # Introduction This opening chapter fixes the scope and expectations for the course. Homological algebra studies algebraic objects by replacing difficult questions about objects with more flexible questions about diagrams, complexes, and resolutions. The course starts from modules and exact sequences, then builds the language needed to discuss derived functors in a subsequent course without developing $\operatorname{Ext}$, $\operatorname{Tor}$, or spectral sequences here. The central theme is that exactness records whether information is lost, created, or transported correctly through a sequence of homomorphisms. Chain complexes organise many exactness questions at once, while resolutions replace an arbitrary module by a long exact algebraic model assembled from projective or injective modules. The purpose of this course is to make these replacements precise and to develop the diagram techniques that make them usable. ## What Homological Algebra Measures A first problem in algebra is that many natural constructions preserve only part of the information in a sequence. Tensoring, taking invariants, applying Hom, and passing to quotients often interact well with some maps but not with all exact sequences. Homological algebra begins by measuring this failure rather than hiding it. [explanation: Exactness As Information Control] A sequence of module homomorphisms records a chain of algebraic operations. Exactness at one term says that everything killed by the outgoing map was already produced by the incoming map. Thus exactness is a bookkeeping device for distinguishing genuine new obstructions from elements that have already been accounted for. This viewpoint appears throughout the course. Homology will measure the discrepancy between cycles and boundaries in a complex. Connecting homomorphisms will explain how a failure at one degree is transported into the next. Resolutions will replace a module by a complex whose failure of exactness is concentrated at a chosen location. [/explanation] This explanation turns a familiar condition into a guiding question: if exactness fails, can we name the failure and move it through a diagram? To make that question precise, all kernels, images, quotients, and exact sequences must live in one category where homomorphisms have compatible domains and codomains. The course therefore fixes the side of the module action before any homological constructions begin. [definition: Module Category Setting] Let $R$ be a ring with multiplicative identity. The course works in the category $R\text{-}\mathrm{Mod}$ of left $R$-modules and $R$-linear maps. [/definition] For commutative rings the distinction between left and right modules is often suppressed, but in homological algebra it is useful to keep sidedness visible. Many constructions, especially tensor products and Hom functors, depend on whether the module action is on the left or on the right. [example: A First Exact Sequence] Let $n \ge 2$. In the category of abelian groups, equivalently $\mathbb Z$-modules, define $i:n\mathbb Z\to \mathbb Z$ by $i(nk)=nk$ and define $q:\mathbb Z\to \mathbb Z/n\mathbb Z$ by $q(m)=m+n\mathbb Z$. We show that \begin{align*} 0 \longrightarrow n\mathbb Z \xrightarrow{i} \mathbb Z \xrightarrow{q} \mathbb Z/n\mathbb Z \longrightarrow 0 \end{align*} is exact by checking equality of image and kernel at each nonzero term. At $n\mathbb Z$, exactness means that $\ker i=\operatorname{im}(0\to n\mathbb Z)$. If $x\in n\mathbb Z$ and $i(x)=0$, then $x=0$ because $i$ is inclusion into $\mathbb Z$. Hence \begin{align*} \ker i=\{0\}=\operatorname{im}(0\to n\mathbb Z). \end{align*} At $\mathbb Z$, exactness means $\operatorname{im} i=\ker q$. The image of $i$ is \begin{align*} \operatorname{im} i=\{i(nk):k\in\mathbb Z\}=\{nk:k\in\mathbb Z\}=n\mathbb Z. \end{align*} For the kernel of $q$, \begin{align*} m\in\ker q &\Longleftrightarrow q(m)=0+n\mathbb Z\\ &\Longleftrightarrow m+n\mathbb Z=0+n\mathbb Z\\ &\Longleftrightarrow m\in n\mathbb Z. \end{align*} Thus $\ker q=n\mathbb Z=\operatorname{im} i$. At $\mathbb Z/n\mathbb Z$, exactness means that $q$ is surjective, since the next map is the zero map to $0$ and its kernel is all of $\mathbb Z/n\mathbb Z$. If $m+n\mathbb Z\in \mathbb Z/n\mathbb Z$, then \begin{align*} q(m)=m+n\mathbb Z, \end{align*} so every coset is in $\operatorname{im}q$. Therefore \begin{align*} \operatorname{im}q=\mathbb Z/n\mathbb Z=\ker(\mathbb Z/n\mathbb Z\to 0). \end{align*} The sequence is exact because, at each term, the elements killed by the outgoing map are exactly the elements produced by the incoming map. [/example] This example is elementary, but it already contains the pattern used later: identify kernels, identify images, and compare them at each position. Diagram lemmas systematise this comparison when several such sequences interact. ## The Objects Built in the Course The next problem is that a single exact sequence is too short to remember repeated algebraic structure. Many constructions naturally produce long strings of maps, and the condition that consecutive maps compose to zero is the minimal compatibility needed for homology. [definition: Chain Complex] A chain complex $C_\bullet$ of $R$-modules is a sequence of $R$-modules and $R$-linear maps \begin{align*} \cdots \xrightarrow{d_{n+2}} C_{n+1} \xrightarrow{d_{n+1}} C_n \xrightarrow{d_n} C_{n-1} \xrightarrow{d_{n-1}} \cdots \end{align*} such that $d_n \circ d_{n+1} = 0$ for every $n \in \mathbb Z$. [/definition] The maps $d_n$ are called differentials. The relation $d_n d_{n+1}=0$ ensures that every boundary is a cycle. The remaining question is how much larger the cycle module is than the boundary module in each degree. That excess is exactly the obstruction to exactness at $C_n$, so it is recorded as a quotient rather than as a raw set of cycles. [definition: Homology Of A Chain Complex] Let $C_\bullet$ be a chain complex of $R$-modules. The $n$-cycles, $n$-boundaries, and $n$-th homology module are \begin{align*} Z_n(C) &= \ker(d_n), & B_n(C) &= \operatorname{im}(d_{n+1}), & H_n(C) &= Z_n(C)/B_n(C). \end{align*} [/definition] Homology is designed to package exactness degree by degree. The following elementary criterion is the first reason cycles and boundaries are the right objects to name. [quotetheorem:4525] [citeproof:4525] This criterion turns exactness into a computable homology statement. The course first develops it carefully, then studies maps between complexes and the induced maps on homology. [example: Multiplication By An Element] Let $R$ be a commutative ring and let $a \in R$. Place the left copy of $R$ in degree $1$, the right copy in degree $0$, and define \begin{align*} d_1:R&\to R, & d_1(r)&=ar,\\ d_0:R&\to 0, & d_0(s)&=0. \end{align*} Then $d_0(d_1(r))=d_0(ar)=0$ for every $r\in R$, so \begin{align*} 0 \longrightarrow R \xrightarrow{\,a\,} R \longrightarrow 0 \end{align*} is a chain complex. In degree $1$, the incoming differential is the zero map $0\to R$, so $B_1=\operatorname{im}(0\to R)=0$. The cycles are \begin{align*} Z_1 &=\ker(d_1)\\ &=\{r\in R:d_1(r)=0\}\\ &=\{r\in R:ar=0\}\\ &=\operatorname{Ann}_R(a). \end{align*} Therefore \begin{align*} H_1=Z_1/B_1=\operatorname{Ann}_R(a)/0\cong \operatorname{Ann}_R(a). \end{align*} In degree $0$, the outgoing differential is $d_0:R\to 0$, so \begin{align*} Z_0=\ker(d_0)=R. \end{align*} The boundaries are the image of multiplication by $a$: \begin{align*} B_0 &=\operatorname{im}(d_1)\\ &=\{d_1(r):r\in R\}\\ &=\{ar:r\in R\}\\ &=aR. \end{align*} Hence \begin{align*} H_0=Z_0/B_0=R/aR. \end{align*} Thus $H_1$ detects elements killed by multiplication by $a$, while $H_0$ records the quotient obtained by identifying all multiples of $a$ with zero. [/example] This small complex shows why homology is not just an abstract quotient. It extracts familiar algebraic data from a structured sequence of maps, and later examples use the same mechanism in longer complexes. ## Diagram Lemmas As Algebraic Transport Once complexes and exact sequences are available, the next problem is comparison. Algebra often gives two rows, two columns, or a commutative square, and the task is to infer the missing kernel, cokernel, or exactness statement from the rest of the diagram. A particularly important warning is that exactness does not by itself mean direct-sum decomposition. The next theorem identifies the extra maps whose existence turns a short exact sequence into a genuine decomposition. [quotetheorem:4526] [citeproof:4526] This theorem is the first model for many later arguments. A structural property of a sequence is translated into the existence of a map, and the proof proceeds by decomposing elements using exactness. The short exactness hypotheses are essential: exactness identifies $A$ with $\ker p$ and ensures that $p$ really has quotient object $C$. Without those conditions, a section or retraction need not control all of $B$. The theorem also does not say that every short exact sequence splits; it says that splitting is equivalent to the existence of one of the additional maps listed above. The next example gives the standard counterexample over $\mathbb Z$. [example: A Non-Split Sequence Over The Integers] The exact sequence \begin{align*} 0 \longrightarrow \mathbb Z \xrightarrow{\,2\,} \mathbb Z \xrightarrow{q} \mathbb Z/2\mathbb Z \longrightarrow 0 \end{align*} does not split in $\mathbb Z\text{-}\mathrm{Mod}$, where $q(m)=m+2\mathbb Z$. We show that $q$ has no $\mathbb Z$-linear section. Suppose, for contradiction, that there is a $\mathbb Z$-linear map $s:\mathbb Z/2\mathbb Z\to \mathbb Z$ such that $q\circ s=\operatorname{id}_{\mathbb Z/2\mathbb Z}$. Let \begin{align*} m=s(1+2\mathbb Z). \end{align*} Since $q\circ s=\operatorname{id}_{\mathbb Z/2\mathbb Z}$, we have \begin{align*} q(m) &=q(s(1+2\mathbb Z))\\ &=(q\circ s)(1+2\mathbb Z)\\ &=1+2\mathbb Z. \end{align*} Thus $m+2\mathbb Z=1+2\mathbb Z$, so $m$ is odd. On the other hand, $1+2\mathbb Z$ has additive order $2$, because \begin{align*} 2(1+2\mathbb Z)=2+2\mathbb Z=0+2\mathbb Z. \end{align*} Using $\mathbb Z$-linearity of $s$, \begin{align*} 0 &=s(0+2\mathbb Z)\\ &=s(2(1+2\mathbb Z))\\ &=2s(1+2\mathbb Z)\\ &=2m. \end{align*} Since $\mathbb Z$ has no nonzero element killed by multiplication by $2$, the equation $2m=0$ forces $m=0$. But $m=0$ is even, contradicting that $m$ is odd. Therefore no section $s$ exists, so the short exact sequence is exact but not split. [/example] The contrast between split and non-split sequences motivates the diagram lemmas: exactness alone is strong, but it does not make every sequence a direct sum. The snake lemma, five lemma, and related tools explain what information remains available without a splitting. ## Homotopy And Resolutions A later problem in the course is that different complexes may encode the same homological information. To compare them efficiently, we need a notion of sameness weaker than equality of maps but strong enough to preserve homology. [definition: Chain Homotopy] Let $f,g:C_\bullet \to D_\bullet$ be maps of chain complexes. A chain homotopy from $f$ to $g$ is a family of $R$-linear maps $h_n:C_n\to D_{n+1}$ such that \begin{align*} f_n-g_n = d_{n+1}^D h_n + h_{n-1}d_n^C \end{align*} for every $n \in \mathbb Z$. [/definition] Chain homotopy is designed so that the difference between $f$ and $g$ becomes invisible after passing to homology. This is the first point where homological algebra behaves like algebraic topology, although all arguments in this course stay inside module categories. The point that must be checked is not merely that $f-g$ has a special formula. If $z$ is a cycle, then the term involving $d_Cz$ disappears, and the remaining term must land inside the boundary subgroup of $D$. This is the obstruction to well-defined equality on homology classes: the homotopy identity has to show that $f(z)$ and $g(z)$ differ by a boundary for every cycle $z$. [quotetheorem:4527] [citeproof:4527] This result illustrates the method of the course: write down the relevant identity, restrict it to cycles, and use the boundary subgroup to identify homology classes. The hypotheses are not cosmetic. If $f$ and $g$ are merely arbitrary graded maps, they may not send cycles to cycles, so there is no induced map on homology. If they are chain maps but not chain homotopic, the conclusion can fail: for a complex with zero differential, the zero map and the identity map are chain maps, but they induce different maps on homology and cannot be related by the displayed homotopy identity. Thus chain homotopy is a controlled weakening of equality, not a statement that all maps with the same source and target have the same effect. The next obstruction is that a module may be too complicated for a functor such as $\operatorname{Hom}$ or tensor product to handle exactly. A resolution replaces that module by a complex built from projective modules, where lifting and comparison arguments are more flexible, while keeping the original module visible at the end. [definition: Projective Resolution] Let $M$ be a left $R$-module. A projective resolution of $M$ is an exact sequence \begin{align*} \cdots \longrightarrow P_2 \longrightarrow P_1 \longrightarrow P_0 \longrightarrow M \longrightarrow 0 \end{align*} where each $P_i$ is a projective left $R$-module. [/definition] The module $M$ may not itself be projective, but a projective resolution expresses it as the homological shadow of projective modules. This is the input for derived functors in later courses: apply a functor to the resolution, then read off the resulting homology. [example: A Free Resolution Of A Cyclic Module] For $n \ge 2$, consider the $\mathbb Z$-linear maps \begin{align*} \mu_n:\mathbb Z&\to \mathbb Z, & \mu_n(k)&=nk,\\ q:\mathbb Z&\to \mathbb Z/n\mathbb Z, & q(m)&=m+n\mathbb Z. \end{align*} We show that \begin{align*} 0 \longrightarrow \mathbb Z \xrightarrow{\ \mu_n\ } \mathbb Z \xrightarrow{\ q\ } \mathbb Z/n\mathbb Z \longrightarrow 0 \end{align*} is exact and hence is a free resolution of $\mathbb Z/n\mathbb Z$. First, $\mu_n$ is injective. Indeed, if $k\in\ker(\mu_n)$, then \begin{align*} \mu_n(k)=0 &\Longleftrightarrow nk=0. \end{align*} Since $n\ne 0$ in the integral domain $\mathbb Z$, this forces $k=0$. Hence \begin{align*} \ker(\mu_n)=\{0\}=\operatorname{im}(0\to\mathbb Z). \end{align*} Next we compare $\operatorname{im}(\mu_n)$ with $\ker(q)$. The image of $\mu_n$ is \begin{align*} \operatorname{im}(\mu_n) &=\{\mu_n(k):k\in\mathbb Z\}\\ &=\{nk:k\in\mathbb Z\}\\ &=n\mathbb Z. \end{align*} For the kernel of $q$, \begin{align*} m\in\ker(q) &\Longleftrightarrow q(m)=0+n\mathbb Z\\ &\Longleftrightarrow m+n\mathbb Z=0+n\mathbb Z\\ &\Longleftrightarrow m-0\in n\mathbb Z\\ &\Longleftrightarrow m\in n\mathbb Z. \end{align*} Therefore \begin{align*} \ker(q)=n\mathbb Z=\operatorname{im}(\mu_n). \end{align*} Finally, $q$ is surjective because every element of $\mathbb Z/n\mathbb Z$ has the form $m+n\mathbb Z$ for some $m\in\mathbb Z$, and \begin{align*} q(m)=m+n\mathbb Z. \end{align*} Thus \begin{align*} \operatorname{im}(q)=\mathbb Z/n\mathbb Z=\ker(\mathbb Z/n\mathbb Z\to 0). \end{align*} Both copies of $\mathbb Z$ are free $\mathbb Z$-modules, so this exact sequence is a free resolution of $\mathbb Z/n\mathbb Z$. It replaces the quotient module by free modules while keeping the quotient information in the final map $q$. [/example] Injective resolutions are dual to projective resolutions and enter when contravariant or left exact functors are studied. This course develops both sides enough to make the later construction of derived functors natural. ## What The Course Does Not Yet Do The final framing problem is boundary-setting. Homological algebra has a large vocabulary, but this first course is about the infrastructure beneath derived functors rather than the derived functors themselves. [remark: Scope Of The First Course] The course builds chain complexes, homology, diagram chasing, homotopy, projective modules, injective modules, and resolutions. It does not develop $\operatorname{Ext}$, $\operatorname{Tor}$, spectral sequences, derived categories, or model-category language. Those topics depend on the machinery introduced here. [/remark] This restriction is deliberate. By keeping the first course focused on complexes and resolutions, every later construction has a concrete algebraic meaning: a functor is applied to a chosen resolution, and the resulting homology measures the obstruction that exactness failed to preserve. The same language later reappears in algebraic topology through singular homology, in group cohomology through resolutions of the trivial module, and in algebraic geometry through sheaf cohomology; this chapter only flags those destinations. [explanation: Expected Working Skills] By the end of the course, students should be able to construct and manipulate chain complexes, identify cycles and boundaries, prove exactness by diagram chasing, use the snake lemma and five lemma, compare maps up to chain homotopy, and build projective and injective resolutions in standard module categories. The course also trains a style of proof. Many arguments are local at a term of a complex, but their consequences are global across a long sequence. Keeping track of indices, signs, kernels, and quotient maps is not clerical detail; it is the main technique. [/explanation] The introduction has now named the central objects and shown why exactness matters. The next chapter turns that motivation into formal language by defining chain complexes, homology, shifts, truncations, and morphisms of complexes. # 1. Chain Complexes and Exactness This chapter fixes the language in which the rest of the course is written. Instead of studying a single module, we study a sequence of modules connected by maps whose consecutive composites vanish; the resulting homology records exactly where the sequence fails to be exact. The chapter moves from the basic definitions of complexes to exact sequences and then to maps between complexes, because the resolution constructions in Chapters 9 and 10, and the derived functors left for the next course, depend on passing information functorially through homology. ## Complexes, Cycles, Boundaries, and Homology What algebraic structure is needed to measure the failure of exactness systematically? A single homomorphism has a kernel and an image, but exactness is a condition comparing the image of one map with the kernel of the next. Chain complexes package all these comparisons into one object. [definition: Chain Complex] Let $R$ be a ring. A chain complex of left $R$-modules is a family of left $R$-modules $(C_n)_{n \in \mathbb Z}$ together with $R$-module homomorphisms $d_n: C_n \to C_{n-1}$ such that \begin{align*} d_{n-1} \circ d_n = 0 \end{align*} for every $n \in \mathbb Z$. [/definition] The maps $d_n$ are called differentials or boundary maps. The condition $d_{n-1}d_n=0$ is the algebraic form of the slogan that a boundary has no boundary, and it forces every element produced by one differential to lie in the kernel of the next. [definition: Cycles and Boundaries] Let $(C_\bullet,d_\bullet)$ be a chain complex of left $R$-modules. The module of $n$-cycles is \begin{align*} Z_n(C) := \ker(d_n: C_n \to C_{n-1}), \end{align*} and the module of $n$-boundaries is \begin{align*} B_n(C) := \operatorname{im}(d_{n+1}: C_{n+1} \to C_n). \end{align*} [/definition] Since $d_n d_{n+1}=0$, we have $B_n(C) \subset Z_n(C)$. Homology is the quotient measuring the gap between cycles that merely vanish under the next differential and cycles that already came from the previous module. [definition: Homology of a Chain Complex] Let $(C_\bullet,d_\bullet)$ be a chain complex of left $R$-modules. The $n$th homology module of $C_\bullet$ is \begin{align*} H_n(C) := Z_n(C) / B_n(C). \end{align*} [/definition] Thus $H_n(C)=0$ exactly when every $n$-cycle is an $n$-boundary. This is the basic local test for exactness of a complex at degree $n$. [example: Multiplication by an Element] Let $R$ be a commutative ring, let $x \in R$, and place $R$ in degrees $1$ and $0$ with differential \begin{align*} d_1: R \to R, \qquad d_1(r)=xr, \end{align*} and with all other differentials zero: \begin{align*} 0 \longrightarrow R \xrightarrow{x} R \longrightarrow 0. \end{align*} In degree $1$ we have \begin{align*} Z_1(C) &= \ker(d_1: R \to R) \\ &= \{r \in R : d_1(r)=0\} \\ &= \{r \in R : xr=0\} \\ &= \operatorname{Ann}_R(x), \end{align*} while \begin{align*} B_1(C) &= \operatorname{im}(d_2:0 \to R) \\ &=0. \end{align*} Therefore \begin{align*} H_1(C) &= Z_1(C)/B_1(C) \\ &= \operatorname{Ann}_R(x)/0 \\ &\cong \operatorname{Ann}_R(x). \end{align*} In degree $0$, the outgoing differential is $d_0:R \to 0$, so \begin{align*} Z_0(C) &= \ker(d_0:R \to 0) \\ &= R, \end{align*} and the incoming boundaries are \begin{align*} B_0(C) &= \operatorname{im}(d_1:R \to R) \\ &= \{d_1(r):r \in R\} \\ &= \{xr:r \in R\} \\ &= xR. \end{align*} Thus \begin{align*} H_0(C) &= Z_0(C)/B_0(C) \\ &= R/xR. \end{align*} The degree $1$ homology detects exactly the elements killed by multiplication by $x$, while the degree $0$ homology records the quotient by the ideal generated by $x$. [/example] This example is the first shadow of the Koszul construction: multiplication maps can be arranged into complexes whose homology detects regularity properties. Topology gives another source of the same language, where modules encode cells and differentials encode incidence data. [example: Cellular Chain Complex of a Circle] Give $S^1$ the CW structure obtained by attaching one $1$-cell $e^1$ to one $0$-cell $e^0$. Then the cellular chain groups are \begin{align*} C_1(S^1) &\cong \mathbb Z\langle e^1\rangle,\\ C_0(S^1) &\cong \mathbb Z\langle e^0\rangle, \end{align*} and $C_n(S^1)=0$ for $n\ne 0,1$. The boundary of the $1$-cell is computed from the two endpoints of the interval model for $e^1$: one endpoint contributes $+e^0$, and the other contributes $-e^0$. Since both endpoints are attached to the same $0$-cell, we get \begin{align*} d_1(e^1) &= e^0-e^0\\ &=0. \end{align*} Thus $d_1:C_1(S^1)\to C_0(S^1)$ is the zero homomorphism. In degree $1$, \begin{align*} Z_1(S^1) &= \ker(d_1:C_1(S^1)\to C_0(S^1))\\ &= \ker(0:\mathbb Z\to \mathbb Z)\\ &= \mathbb Z, \end{align*} while \begin{align*} B_1(S^1) &= \operatorname{im}(d_2:C_2(S^1)\to C_1(S^1))\\ &= \operatorname{im}(0:0\to \mathbb Z)\\ &=0. \end{align*} Therefore \begin{align*} H_1(S^1) &= Z_1(S^1)/B_1(S^1)\\ &= \mathbb Z/0\\ &\cong \mathbb Z. \end{align*} In degree $0$, the outgoing differential is $d_0:C_0(S^1)\to C_{-1}(S^1)=0$, so \begin{align*} Z_0(S^1) &= \ker(d_0:C_0(S^1)\to 0)\\ &= \mathbb Z, \end{align*} and \begin{align*} B_0(S^1) &= \operatorname{im}(d_1:C_1(S^1)\to C_0(S^1))\\ &= \operatorname{im}(0:\mathbb Z\to \mathbb Z)\\ &=0. \end{align*} Hence \begin{align*} H_0(S^1) &= Z_0(S^1)/B_0(S^1)\\ &= \mathbb Z/0\\ &\cong \mathbb Z. \end{align*} The nonzero class in degree $1$ records the loop around the circle, and the copy of $\mathbb Z$ in degree $0$ records that the CW complex has one connected component. [/example] The circle computation used a chain complex, where boundaries lower degree. Cohomological constructions force the opposite bookkeeping problem: maps such as those in Hom-complexes naturally send degree $n$ data to degree $n+1$ data. To use the same condition that two consecutive differentials compose to zero without reversing every arrow by hand, we need the upward-degree version of a complex. [definition: Cochain Complex] Let $R$ be a ring. A cochain complex of left $R$-modules is a family of left $R$-modules $(C^n)_{n \in \mathbb Z}$ together with $R$-module homomorphisms $d^n: C^n \to C^{n+1}$ such that \begin{align*} d^{n+1} \circ d^n = 0 \end{align*} for every $n \in \mathbb Z$. [/definition] The cochain convention is useful when maps naturally point upward in degree, as with cohomology theories and Hom-complexes. Once the differential raises degree, the same exactness question must be asked with shifted indices: which elements are killed by the outgoing map, and which have already arrived from the previous degree? This produces cocycles, coboundaries, and cohomology in direct analogy with the chain case. [definition: Cohomology of a Cochain Complex] Let $(C^\bullet,d^\bullet)$ be a cochain complex of left $R$-modules. The module of $n$-cocycles is \begin{align*} Z^n(C) := \ker(d^n: C^n \to C^{n+1}), \end{align*} the module of $n$-coboundaries is \begin{align*} B^n(C) := \operatorname{im}(d^{n-1}: C^{n-1} \to C^n), \end{align*} and the $n$th cohomology module is \begin{align*} H^n(C) := Z^n(C) / B^n(C). \end{align*} [/definition] Complexes are often reindexed without changing their mathematical content. In long exact sequences, mapping cones, and derived constructions, the same module may need to be viewed as living in a different degree. A shift is the controlled operation that changes the labels while adjusting signs so the differential still squares to zero. [definition: Shift of a Chain Complex] Let $C_\bullet$ be a chain complex. For $k \in \mathbb Z$, the shifted chain complex $C[k]_\bullet$ is defined by \begin{align*} C[k]_n := C_{n-k}, \end{align*} with differential $d^{C[k]}_n := (-1)^k d^C_{n-k}$. [/definition] The sign convention makes shifts compatible with later constructions on Hom-complexes and mapping cones. In a first reading, the main point is that shifting changes degree labels while preserving the pattern of homology up to the same reindexing. A different operation is needed when only part of a complex is relevant. To study bounded ranges, finite pieces of resolutions, or degrees below a chosen cutoff, one discards modules outside the range and keeps the surviving differentials where they still make sense. [definition: Brutal Truncation of a Chain Complex] Let $C_\bullet$ be a chain complex and let $m \in \mathbb Z$. The brutal truncation $\tau_{\le m}C$ is the chain complex with \begin{align*} (\tau_{\le m}C)_n := \begin{cases} C_n, & n \le m,\\ 0, & n > m, \end{cases} \end{align*} and differentials induced from $C_\bullet$ wherever both source and target are nonzero. [/definition] Truncation is a way of isolating a range of degrees. Later it will let us compare finite pieces of resolutions and control where homology can occur. ## Exactness as Vanishing Homology When should a complex be regarded as having no new information at a given degree? The answer is that all cycles should already be boundaries. This turns exactness, usually phrased term-by-term, into the vanishing of homology. [definition: Exactness at a Term] A sequence of $R$-module homomorphisms \begin{align*} A \xrightarrow{f} B \xrightarrow{g} C \end{align*} is exact at $B$ if $\operatorname{im} f = \ker g$. [/definition] This definition is local: it only involves the module in the middle and the two adjacent maps. A chain complex is exact when this local condition holds everywhere. [definition: Exact Complex] A chain complex $C_\bullet$ is exact if \begin{align*} \operatorname{im}(d_{n+1}) = \ker(d_n) \end{align*} for every $n \in \mathbb Z$. [/definition] The definition of an exact complex says that there is no leftover cycle at any degree: everything killed by the outgoing differential must already have arrived from the incoming differential. The homology modules were introduced precisely to measure this leftover part, degree by degree. Thus the next result is not an additional structure theorem, but the formal check that the quotient construction has captured exactly the usual notion of exactness. [quotetheorem:4528] [citeproof:4528] This theorem explains why homological algebra treats exact complexes as acyclic objects: they contain no homology. Its scope is deliberately degreewise. Vanishing homology says that each cycle is locally a boundary, but it does not say that the complex decomposes into direct sums, that the differentials have chosen splittings, or that the complex is contractible. Those stronger conclusions require extra structure, and the first place where this distinction matters is the theory of short exact sequences. [definition: Short Exact Sequence] A short exact sequence of left $R$-modules is an exact sequence \begin{align*} 0 \longrightarrow A \xrightarrow{i} B \xrightarrow{p} C \longrightarrow 0. \end{align*} [/definition] Exactness at $A$ says that $i$ is injective, exactness at $B$ says that $\operatorname{im} i = \ker p$, and exactness at $C$ says that $p$ is surjective. Thus $B$ contains a copy of $A$, and $C$ is the corresponding quotient $B/i(A)$. A useful way to read a short exact sequence is therefore as a controlled presentation of one module as an extension of another: the left term is the submodule, the right term is the quotient, and the middle term records how the two are glued together. [example: Integers Modulo n] Let $n \ge 1$, let $i:n\mathbb Z \to \mathbb Z$ be the inclusion $i(nk)=nk$, and let $q:\mathbb Z \to \mathbb Z/n\mathbb Z$ be the quotient map $q(a)=\bar a$. We show that \begin{align*} 0 \longrightarrow n\mathbb Z \xrightarrow{i} \mathbb Z \xrightarrow{q} \mathbb Z/n\mathbb Z \longrightarrow 0 \end{align*} is exact by checking the three middle terms. First, $i$ is injective: if $i(nk)=0$, then $nk=0$ in $\mathbb Z$, so $k=0$ because $n\ge 1$, and therefore $nk=0$ in $n\mathbb Z$. Thus \begin{align*} \ker i=0. \end{align*} Next, compute the kernel of $q$: \begin{align*} \ker q &= \{a\in \mathbb Z : q(a)=\bar 0\}\\ &= \{a\in \mathbb Z : \bar a=\bar 0 \text{ in } \mathbb Z/n\mathbb Z\}\\ &= \{a\in \mathbb Z : n \text{ divides } a\}\\ &= \{nk : k\in \mathbb Z\}\\ &= n\mathbb Z. \end{align*} Since $i(nk)=nk$, the image of $i$ is \begin{align*} \operatorname{im} i &= \{i(nk):k\in \mathbb Z\}\\ &= \{nk:k\in \mathbb Z\}\\ &= n\mathbb Z. \end{align*} Hence $\operatorname{im} i=\ker q$, so the sequence is exact at $\mathbb Z$. Finally, $q$ is surjective: every element of $\mathbb Z/n\mathbb Z$ has the form $\bar a$ for some $a\in \mathbb Z$, and $q(a)=\bar a$. Therefore \begin{align*} \operatorname{im} q=\mathbb Z/n\mathbb Z. \end{align*} The sequence is exact, and it realizes $\mathbb Z/n\mathbb Z$ as the quotient of $\mathbb Z$ by the submodule $n\mathbb Z$. [/example] The quotient example shows how a short exact sequence records a submodule and its quotient, but it also exposes a missing question: does the middle module contain the quotient as an independent $R$-linear piece, or is the quotient only visible after collapsing the submodule? To name the sequences with no hidden extension data, we isolate the case where the middle term is compatibly a direct sum of the two end terms. [definition: Split Short Exact Sequence] A short exact sequence \begin{align*} 0 \longrightarrow A \xrightarrow{i} B \xrightarrow{p} C \longrightarrow 0 \end{align*} is split if there exists an $R$-module isomorphism $\varphi: B \to A \oplus C$ such that $\varphi \circ i(a)=(a,0)$ for every $a \in A$ and $p \circ \varphi^{-1}(a,c)=c$ for every $(a,c) \in A \oplus C$. [/definition] The definition gives splitting as a compatible direct sum decomposition of the middle term, but in practice one rarely starts with the isomorphism $B \cong A \oplus C$ already written down. There are two more usable ways to detect the same structure: split the quotient map by choosing representatives $R$-linearly, or split the inclusion by projecting back onto the submodule. The point of the next theorem is that either one-sided piece of extra structure forces the whole short exact sequence to be a direct sum extension. [quotetheorem:4529] [citeproof:4529] This criterion separates exactness from splitting. Exactness says $C$ is the quotient $B/A$; splitting says the quotient has been lifted back into $B$ by an $R$-linear choice. Such a choice is real extra structure, and it can fail even for very small abelian groups. [example: Nonsplit Extension of Z Mod Two] Let $\overline{a}_m$ denote the class of $a$ in $\mathbb Z/m\mathbb Z$. Consider \begin{align*} 0 \longrightarrow \mathbb Z/2\mathbb Z \xrightarrow{i} \mathbb Z/4\mathbb Z \xrightarrow{p} \mathbb Z/2\mathbb Z \longrightarrow 0, \end{align*} where $i(\overline{1}_2)=\overline{2}_4$ and $p(\overline{a}_4)=\overline{a}_2$. Since $\mathbb Z/2\mathbb Z=\{\overline{0}_2,\overline{1}_2\}$, we have \begin{align*} i(\overline{0}_2)&=\overline{0}_4,\\ i(\overline{1}_2)&=\overline{2}_4\ne \overline{0}_4, \end{align*} so $\ker i=0$. The image of $i$ is \begin{align*} \operatorname{im} i &=\{i(\overline{0}_2),i(\overline{1}_2)\}\\ &=\{\overline{0}_4,\overline{2}_4\}. \end{align*} Also, \begin{align*} \ker p &=\{\overline{a}_4\in \mathbb Z/4\mathbb Z:p(\overline{a}_4)=\overline{0}_2\}\\ &=\{\overline{a}_4:a\equiv 0 \pmod 2\}\\ &=\{\overline{0}_4,\overline{2}_4\}, \end{align*} so $\operatorname{im} i=\ker p$. Finally, \begin{align*} p(\overline{0}_4)&=\overline{0}_2,\\ p(\overline{1}_4)&=\overline{1}_2, \end{align*} so $p$ is surjective. Thus the sequence is exact. It remains to show that it is not split. By *Splitting Criteria for Short Exact Sequences*, a split short exact sequence would admit a section $s:\mathbb Z/2\mathbb Z\to \mathbb Z/4\mathbb Z$ with $p\circ s=\operatorname{id}_{\mathbb Z/2\mathbb Z}$. If such an $s$ existed, then \begin{align*} p(s(\overline{1}_2))=\overline{1}_2, \end{align*} so $s(\overline{1}_2)$ would have to be one of the two classes mapping to $\overline{1}_2$: \begin{align*} s(\overline{1}_2)\in \{\overline{1}_4,\overline{3}_4\}. \end{align*} But $s$ is a group homomorphism, and $2\overline{1}_2=\overline{0}_2$, so \begin{align*} 2s(\overline{1}_2) &=s(2\overline{1}_2)\\ &=s(\overline{0}_2)\\ &=\overline{0}_4. \end{align*} This is impossible, because \begin{align*} 2\overline{1}_4&=\overline{2}_4\ne \overline{0}_4,\\ 2\overline{3}_4&=\overline{6}_4=\overline{2}_4\ne \overline{0}_4. \end{align*} Therefore no section exists, so the exact sequence is not split. This example shows that exactness can identify the quotient without providing an internal direct-sum decomposition of the middle term. [/example] The nonsplit example shows why the splitting criteria are useful: exactness alone identifies the quotient, while splitting asks whether the quotient can be represented inside the middle term without distorting the module structure. The split model below is the contrasting case where the required section and retraction are visible. [example: Noncanonical Splitting over the Integers] Let \begin{align*} i:\mathbb Z \to \mathbb Z\oplus \mathbb Z/2\mathbb Z,\qquad i(a)=(a,\bar 0), \end{align*} and \begin{align*} p:\mathbb Z\oplus \mathbb Z/2\mathbb Z \to \mathbb Z/2\mathbb Z,\qquad p(a,\bar b)=\bar b. \end{align*} The sequence \begin{align*} 0 \longrightarrow \mathbb Z \xrightarrow{i} \mathbb Z \oplus \mathbb Z/2\mathbb Z \xrightarrow{p} \mathbb Z/2\mathbb Z \longrightarrow 0 \end{align*} is exact and split. First, $i$ is injective. If $i(a)=(0,\bar 0)$, then \begin{align*} (a,\bar 0)=(0,\bar 0), \end{align*} so $a=0$. Hence $\ker i=0$. Next, \begin{align*} \ker p &=\{(a,\bar b)\in \mathbb Z\oplus \mathbb Z/2\mathbb Z : p(a,\bar b)=\bar 0\}\\ &=\{(a,\bar b)\in \mathbb Z\oplus \mathbb Z/2\mathbb Z : \bar b=\bar 0\}\\ &=\{(a,\bar 0):a\in \mathbb Z\}. \end{align*} On the other hand, \begin{align*} \operatorname{im} i &=\{i(a):a\in \mathbb Z\}\\ &=\{(a,\bar 0):a\in \mathbb Z\}. \end{align*} Therefore $\operatorname{im} i=\ker p$, so the sequence is exact at $\mathbb Z\oplus \mathbb Z/2\mathbb Z$. Finally, $p$ is surjective because for every $\bar b\in \mathbb Z/2\mathbb Z$, \begin{align*} p(0,\bar b)=\bar b. \end{align*} Thus the sequence is short exact. To split it, define \begin{align*} s:\mathbb Z/2\mathbb Z \to \mathbb Z\oplus \mathbb Z/2\mathbb Z,\qquad s(\bar b)=(0,\bar b). \end{align*} Then \begin{align*} (p\circ s)(\bar b) &=p(s(\bar b))\\ &=p(0,\bar b)\\ &=\bar b, \end{align*} so $p\circ s=\operatorname{id}_{\mathbb Z/2\mathbb Z}$. By *Splitting Criteria for Short Exact Sequences*, the short exact sequence is split. Equivalently, the retraction \begin{align*} r:\mathbb Z\oplus \mathbb Z/2\mathbb Z \to \mathbb Z,\qquad r(a,\bar b)=a \end{align*} satisfies \begin{align*} (r\circ i)(a) &=r(i(a))\\ &=r(a,\bar 0)\\ &=a, \end{align*} so $r\circ i=\operatorname{id}_{\mathbb Z}$. This example realizes the middle term as the direct sum of the embedded copy of $\mathbb Z$ and the quotient $\mathbb Z/2\mathbb Z$, with the splitting determined by the chosen section $s$. [/example] ## Maps of Complexes and Functoriality of Homology How does homology behave under a map from one complex to another? Since homology is built from kernels, images, and quotients in each degree, a degreewise family of module maps must respect the differentials before it can descend to homology. [definition: Chain Map] Let $(C_\bullet,d^C_\bullet)$ and $(D_\bullet,d^D_\bullet)$ be chain complexes of left $R$-modules. A chain map $f: C_\bullet \to D_\bullet$ is a family of $R$-module homomorphisms $f_n: C_n \to D_n$ such that \begin{align*} d^D_n \circ f_n = f_{n-1} \circ d^C_n \end{align*} for every $n \in \mathbb Z$. [/definition] This commutation relation says that applying $f$ and then taking a boundary agrees with taking a boundary and then applying $f$. The next issue is whether a chain map produces a well-defined map after quotienting cycles by boundaries. It must carry cycles to cycles and boundaries to boundaries, so that changing a representative by a boundary does not change the resulting homology class. [quotetheorem:1922] [citeproof:1922] The chain map hypothesis is essential here. A degreewise family of maps that does not commute with the differentials need not send cycles to cycles, and even if it happens to send a particular cycle to a cycle, it may send homologous representatives to non-homologous elements. The commutative square condition is exactly what makes the formula on homology independent of representatives. Once that condition is imposed, the construction is compatible with identity maps and composition. [quotetheorem:4530] [citeproof:4530] Functoriality gives the formal mechanism behind later diagram lemmas. Once a construction is made at the level of complexes and checked to commute with differentials, it automatically produces maps on homology. It does not say that induced maps preserve all information: a nonzero chain map can induce the zero map on homology, and functoriality by itself does not assert that applying an arbitrary functor to modules preserves exactness. In computations, the reusable procedure is: identify cycles, identify boundaries, form the quotient, verify chain-map compatibility, and then compute the induced map on representatives. [example: Induced Map for a Multiplication Complex] Let $C_\bullet$ be the two-term complex $R \xrightarrow{x} R$, so $d^C_1(r)=xr$, and let $D_\bullet$ be the two-term complex $R \xrightarrow{y} R$, so $d^D_1(r)=yr$. Define \begin{align*} f_1:R\to R,\qquad f_1(r)=ar, \end{align*} and \begin{align*} f_0:R\to R,\qquad f_0(r)=ar. \end{align*} For $r\in R$, \begin{align*} (d^D_1\circ f_1)(r) &=d^D_1(ar)\\ &=y(ar)\\ &=(ya)r\\ &=(ax)r\\ &=a(xr)\\ &=f_0(xr)\\ &=(f_0\circ d^C_1)(r), \end{align*} where we used the hypothesis $ax=ya$ and commutativity of $R$. In degree $0$ all outgoing differentials are zero, and in all other degrees the modules are zero, so these maps form a chain map. By *Chain Maps Induce Maps on Homology*, the induced map on $H_0$ sends the class of $r$ to the class of $f_0(r)=ar$. Using the earlier computation for a multiplication complex, \begin{align*} H_0(C)&\cong R/xR,\\ H_0(D)&\cong R/yR. \end{align*} Thus the degree $0$ induced map is \begin{align*} R/xR&\longrightarrow R/yR,\\ \bar r&\longmapsto \overline{ar}. \end{align*} This formula is independent of the chosen representative: if $r'=r+xs$, then \begin{align*} ar' &=a(r+xs)\\ &=ar+axs\\ &=ar+yas\\ &=ar+y(as), \end{align*} so $ar'$ and $ar$ differ by an element of $yR$. In degree $1$, \begin{align*} H_1(C)&\cong \operatorname{Ann}_R(x),\\ H_1(D)&\cong \operatorname{Ann}_R(y). \end{align*} If $u\in \operatorname{Ann}_R(x)$, then $xu=0$, and \begin{align*} y(au) &=(ya)u\\ &=(ax)u\\ &=a(xu)\\ &=a\cdot 0\\ &=0. \end{align*} Hence $au\in \operatorname{Ann}_R(y)$, and the induced degree $1$ map is \begin{align*} \operatorname{Ann}_R(x)&\longrightarrow \operatorname{Ann}_R(y),\\ u&\longmapsto au. \end{align*} The condition $ax=ya$ is exactly what makes multiplication by $a$ commute with the two differentials, so multiplication by $a$ descends to the stated maps on homology. [/example] The chapter ends with the guiding principle for the next stage: exactness is homology vanishing, and chain maps are the morphisms that preserve the structure needed to pass to homology. Diagram chasing will refine this principle into tools such as the snake lemma, where exactness information is transported across a commutative diagram. Once chain complexes are in place, exactness becomes something we can test and transport across maps. The next chapter makes that movement explicit through diagram chasing, where the snake lemma turns commutative squares and short exact rows into new exact sequences. # 2. Diagram Chasing and the Snake Lemma Diagram chasing turns exactness from a static condition into a method for constructing maps. In the previous chapter, exact sequences and chain maps recorded when images agree with kernels; here we learn how a commutative diagram forces new exact sequences between kernels and cokernels. The main result is the snake lemma, whose connecting homomorphism is the first instance of a boundary map built by lifting, applying a map, and descending to a quotient. ## Kernels, Cokernels, Images, and Coimages When a diagram contains several rows and columns, the first problem is to track what information is attached to a single module homomorphism. Kernels measure where a map loses injectivity, cokernels measure where it fails to be surjective, and images and coimages express the two quotient constructions that meet in the first isomorphism theorem. [definition: Kernel] Let $R$ be a ring and let $f: M \to N$ be an $R$-module homomorphism. The kernel of $f$ is the submodule \begin{align*} \ker f = \{m \in M : f(m)=0\} \subset M. \end{align*} [/definition] The kernel is the largest part of the domain killed by $f$. It also has a universal property: any map into $M$ whose composite with $f$ is zero factors uniquely through the inclusion $\ker f \hookrightarrow M$. There is a dual failure to measure on the target side. Even when a map is injective, it may miss elements of its codomain, and the quotient by the image records exactly what remains after everything hit by $f$ has been identified with zero. [definition: Cokernel] Let $R$ be a ring and let $f: M \to N$ be an $R$-module homomorphism. The cokernel of $f$ is the quotient module \begin{align*} \operatorname{coker} f = N / \operatorname{im} f. \end{align*} [/definition] The cokernel is the universal target on which $f$ becomes zero. If $q: N \to \operatorname{coker} f$ is the quotient map, then $qf=0$, and every map $g: N \to P$ with $gf=0$ factors uniquely through $q$. Kernels and cokernels describe loss at the source and target, but diagram chasing also needs the actual part of the codomain reached by the map and the corresponding quotient of the domain after the kernel has been collapsed. These two constructions give the image and coimage. [definition: Image and Coimage] Let $R$ be a ring and let $f: M \to N$ be an $R$-module homomorphism. The image of $f$ is \begin{align*} \operatorname{im} f = f(M) \subset N. \end{align*} The coimage of $f$ is \begin{align*} \operatorname{coim} f = M / \ker f. \end{align*} [/definition] The image is a submodule of the codomain, while the coimage is a quotient of the domain. For modules these two objects are identified by the natural map $m+\ker f\mapsto f(m)$, which is the algebraic fact that makes elementwise diagram chasing agree with quotient calculations. [explanation: First Isomorphism Theorem For Modules] For an $R$-module homomorphism $f:M\to N$, the map \begin{align*} M/\ker f &\longrightarrow \operatorname{im}f,\\ m+\ker f &\longmapsto f(m) \end{align*} is a well-defined isomorphism of $R$-modules. [/explanation] This result is used constantly when moving between quotients and submodules. It says that the part of $M$ visible through $f$ can be recorded either as the quotient $M/\ker f$ or as the submodule $\operatorname{im}f\subset N$. The next definition uses this agreement locally: exactness at a term is precisely the statement that the part arriving from the left is the part killed on the right. [definition: Exactness at a Module] Let \begin{align*} A \xrightarrow{f} B \xrightarrow{g} C \end{align*} be homomorphisms of $R$-modules. The sequence is exact at $B$ if \begin{align*} \operatorname{im} f = \ker g. \end{align*} [/definition] Exactness at $B$ includes two assertions: $gf=0$, and every element killed by $g$ comes from $A$. Diagram chases usually prove the second assertion by starting with an arbitrary element in a kernel and modifying it until it is visibly an image. [example: Exactness in a Quotient Sequence] Let $M$ be an $R$-module and let $N \subset M$ be a submodule. Write $i:N\to M$ for the inclusion $i(n)=n$, and write $q:M\to M/N$ for the quotient map $q(m)=m+N$. We verify exactness of \begin{align*} 0 \longrightarrow N \xrightarrow{i} M \xrightarrow{q} M/N \longrightarrow 0. \end{align*} At $N$, the map $0\to N$ has image $\{0\}$, while \begin{align*} \ker i &=\{n\in N : i(n)=0\}\\ &=\{n\in N : n=0\}\\ &=\{0\}. \end{align*} Thus the sequence is exact at $N$. At $M$, first \begin{align*} q(i(n))=q(n)=n+N=N \end{align*} for every $n\in N$, so $\operatorname{im} i\subseteq \ker q$. Conversely, if $m\in \ker q$, then \begin{align*} q(m)=0_{M/N} &\iff m+N=N\\ &\iff m\in N. \end{align*} Hence $\ker q=N=\operatorname{im} i$, so the sequence is exact at $M$. At $M/N$, every element has the form $m+N$ for some $m\in M$, and \begin{align*} q(m)=m+N. \end{align*} Therefore $\operatorname{im}q=M/N$, which is the kernel of the zero map $M/N\to 0$. The sequence is exact at every term, so it is a short exact sequence. [/example] The next diagrams will have exact rows. Before stating the snake lemma, we fix the kind of commutative diagram to which it applies. [definition: Morphism of Short Exact Sequences] A morphism of short exact sequences of $R$-modules is a commutative diagram \begin{align*} 0 \longrightarrow A_1 \xrightarrow{i_1} B_1 \xrightarrow{p_1} C_1 \longrightarrow 0 \\ \phantom{0 \longrightarrow{}} \downarrow^{\alpha} \phantom{A_1} \quad \downarrow^{\beta} \phantom{B_1} \quad \downarrow^{\gamma} \phantom{C_1} \\ 0 \longrightarrow A_0 \xrightarrow{i_0} B_0 \xrightarrow{p_0} C_0 \longrightarrow 0 \end{align*} in which both rows are exact. [/definition] The commutativity conditions are $\beta i_1=i_0\alpha$ and $p_0\beta=\gamma p_1$. The snake lemma extracts from this square of short exact sequences a six-term exact sequence, with a new map from $\ker \gamma$ to $\operatorname{coker}\alpha$. ## The Connecting Homomorphism The essential question in the snake lemma is this: if $c_1 \in C_1$ maps to zero in $C_0$, why should it determine a class in $A_0/\operatorname{im}\alpha$? The answer is forced by exactness. Lift $c_1$ to $B_1$, apply $\beta$, use exactness in the bottom row to move into $A_0$, and then quotient by the ambiguity introduced by the choice of lift. [definition: Connecting Homomorphism] Let \begin{align*} 0 \longrightarrow A_1 \xrightarrow{i_1} B_1 \xrightarrow{p_1} C_1 \longrightarrow 0 \\ \phantom{0 \longrightarrow{}} \downarrow^{\alpha} \phantom{A_1} \quad \downarrow^{\beta} \phantom{B_1} \quad \downarrow^{\gamma} \phantom{C_1} \\ 0 \longrightarrow A_0 \xrightarrow{i_0} B_0 \xrightarrow{p_0} C_0 \longrightarrow 0 \end{align*} be a morphism of short exact sequences. The connecting homomorphism is the map \begin{align*} \delta: \ker\gamma \longrightarrow \operatorname{coker}\alpha \end{align*} defined as follows. For $c_1 \in \ker\gamma$, choose $b_1 \in B_1$ with $p_1(b_1)=c_1$. Since $p_0\beta(b_1)=\gamma p_1(b_1)=0$, choose the unique $a_0 \in A_0$ with $i_0(a_0)=\beta(b_1)$. Define \begin{align*} \delta(c_1) = a_0 + \operatorname{im}\alpha \in \operatorname{coker}\alpha. \end{align*} [/definition] The definition has two possible sources of ambiguity: the lift $b_1$ and the representative of the class in $\operatorname{coker}\alpha$. To use this construction as a map rather than as notation, we must know that both choices disappear in the quotient; the theorem below proves precisely that well-definedness. [quotetheorem:4531] [citeproof:4531] This connecting map is the new ingredient, and its definition explains why every hypothesis in the diagram is doing work. Surjectivity of $p_1$ is what permits the first lift; exactness at $B_0$ is what turns the equation $p_0\beta(b_1)=0$ into an element of $A_0$; injectivity of $i_0$ is what makes that element unique. The quotient by $\operatorname{im}\alpha$ is not decorative: changing the lift by an element of $\operatorname{im}i_1$ changes the resulting element of $A_0$ by exactly something coming from $A_1$. What the construction does not provide is a preferred representative in $A_0$, nor a lift $b_1$ chosen functorially in general; only the class in $\operatorname{coker}\alpha$ is canonical. All other maps in the snake lemma are induced by restricting $\alpha,\beta,\gamma$ to kernels or passing them to cokernels. [example: Boundary Class from a Lift] Take the short exact sequence of abelian groups \begin{align*} 0 \longrightarrow \mathbb Z \xrightarrow{i} \mathbb Q \xrightarrow{p} \mathbb Q/\mathbb Z \longrightarrow 0, \end{align*} where $i(a)=a$ and $p(q)=q+\mathbb Z$, and map this sequence to itself by multiplication by $n$. The left vertical map has image $n\mathbb Z\subset \mathbb Z$, so \begin{align*} \operatorname{coker}(n:\mathbb Z\to \mathbb Z) = \mathbb Z/n\mathbb Z. \end{align*} If $\bar q\in \mathbb Q/\mathbb Z$ lies in the kernel of multiplication by $n$, choose $q\in\mathbb Q$ with $p(q)=\bar q$. Since $n\bar q=0$ in $\mathbb Q/\mathbb Z$, we have \begin{align*} 0 &=n\bar q\\ &=n(q+\mathbb Z)\\ &=nq+\mathbb Z. \end{align*} The zero element of $\mathbb Q/\mathbb Z$ is $\mathbb Z$, so $nq+\mathbb Z=\mathbb Z$, which is equivalent to $nq\in\mathbb Z$. The connecting construction lifts $\bar q$ to $q\in\mathbb Q$, applies the middle vertical map $q\mapsto nq$, and then identifies $nq\in\mathbb Z$ through the inclusion $i:\mathbb Z\to\mathbb Q$. Therefore \begin{align*} \delta(\bar q) = nq+n\mathbb Z \in \mathbb Z/n\mathbb Z. \end{align*} For the specific element $\bar q=\frac{1}{n}+\mathbb Z$, choose the lift $q=\frac{1}{n}$. Then \begin{align*} nq &=n\cdot \frac{1}{n}\\ &=1, \end{align*} so \begin{align*} \delta\left(\frac{1}{n}+\mathbb Z\right) = 1+n\mathbb Z. \end{align*} Thus this boundary class records the integer obstruction produced by multiplying a rational lift by $n$. [/example] This example shows the typical behaviour of $\delta$: torsion in the quotient is measured by an obstruction in the left-hand term. The obstruction vanishes exactly when the chosen lift can be adjusted so that it lies in the kernel of the middle vertical map. ## The Snake Lemma The obstacle is that kernels compare the three vertical maps at the top of the diagram, while cokernels compare the same maps at the bottom. A direct map from $\ker\gamma$ to $\operatorname{coker}\alpha$ is not visible among the original arrows, so the question is whether the lifting construction above supplies exactly the missing bridge. The snake lemma says that it does: the sequence starts with kernels, crosses through the connecting homomorphism, and continues through cokernels. [illustration:homological-algebra-i-snake-lemma-path] [explanation: Snake Lemma] For a morphism of short exact sequences \begin{align*} 0 \longrightarrow A_1 \xrightarrow{i_1} B_1 \xrightarrow{p_1} C_1 \longrightarrow 0 \\ \phantom{0 \longrightarrow{}} \downarrow^{\alpha} \phantom{A_1} \quad \downarrow^{\beta} \phantom{B_1} \quad \downarrow^{\gamma} \phantom{C_1} \\ 0 \longrightarrow A_0 \xrightarrow{i_0} B_0 \xrightarrow{p_0} C_0 \longrightarrow 0, \end{align*} there is an exact sequence \begin{align*} 0\longrightarrow \ker\alpha\longrightarrow \ker\beta\longrightarrow \ker\gamma \xrightarrow{\delta} \operatorname{coker}\alpha\longrightarrow \operatorname{coker}\beta\longrightarrow \operatorname{coker}\gamma\longrightarrow 0, \end{align*} where $\delta$ is the connecting homomorphism defined by lifting through $p_1$, applying $\beta$, and descending through $i_0$. [/explanation] The construction is a model diagram chase: choose representatives, apply commutativity, use exactness to replace a zero condition by membership in a previous image, and then check independence modulo an image. The short exactness hypotheses are not technical decoration; they are what make the lift, the descent, and the endpoint zeros legitimate. [example: Failure Without Surjectivity] Consider the diagram of abelian groups \begin{align*} 0 \longrightarrow 0 \longrightarrow \mathbb Z \xrightarrow{\times 2} \mathbb Z \longrightarrow 0 \\ \phantom{0 \longrightarrow{}} \downarrow \phantom{0} \quad \downarrow^{\operatorname{id}} \phantom{\mathbb Z} \quad \downarrow^{0} \phantom{\mathbb Z} \\ 0 \longrightarrow 0 \longrightarrow \mathbb Z \xrightarrow{0} \mathbb Z \longrightarrow 0. \end{align*} The middle square commutes because, for every $z\in \mathbb Z$, \begin{align*} 0\big((\times 2)(z)\big) &=0(2z)\\ &=0, \end{align*} while \begin{align*} 0\big(\operatorname{id}(z)\big) &=0(z)\\ &=0. \end{align*} Thus the two composites $\mathbb Z\to \mathbb Z$ through that square are equal. The top row is not exact at the right-hand copy of $\mathbb Z$. The image of multiplication by $2$ is \begin{align*} \operatorname{im}(\times 2) &=\{2z:z\in \mathbb Z\}\\ &=2\mathbb Z, \end{align*} whereas the kernel of the zero map $\mathbb Z\to 0$ is \begin{align*} \ker(\mathbb Z\to 0) &=\{z\in \mathbb Z:z\mapsto 0\}\\ &=\mathbb Z. \end{align*} Since $1\in \mathbb Z$ but $1\notin 2\mathbb Z$, the equality $\operatorname{im}(\times 2)=\ker(\mathbb Z\to 0)$ fails. This failure is exactly where the connecting construction breaks. The element $1$ lies in the right-hand vertical kernel because \begin{align*} 0(1)=0, \end{align*} so $1\in \ker(0:\mathbb Z\to \mathbb Z)$. To begin the boundary construction one would need an integer $z$ with \begin{align*} 2z=1. \end{align*} No such $z\in \mathbb Z$ exists, because $2z$ is even for every integer $z$, while $1$ is not even. Therefore the lift required to define the connecting homomorphism does not exist, and the snake sequence cannot be recovered from commutativity alone. [/example] This example isolates the limitation: the lemma is not a statement about arbitrary commutative rectangles. It is a statement about rectangles whose rows are short exact sequences, so that every representative chosen in the chase exists for a reason and every ambiguity is controlled by a named image. [remark: Endpoint Zeros] The zero at the left of the snake sequence uses injectivity of $i_1$ in the top short exact sequence. The zero at the right uses surjectivity of $p_0$ in the bottom short exact sequence. If the diagram has only exact rows without these endpoint hypotheses, a shorter exact sequence remains, but the endpoint zeros may disappear. [/remark] The snake lemma produces a connecting homomorphism from choices of representatives and lifts, so a new problem appears when two snake diagrams are compared at once. A construction useful in later arguments must respect maps between diagrams: carrying a class across the comparison and then applying the boundary should agree with first applying the boundary and then carrying the result across. [quotetheorem:4532] [citeproof:4532] Naturality is often what permits the snake lemma to be used inside larger constructions. It guarantees that the boundary map is not an arbitrary choice from a chase, but a canonical map attached to the original morphism of short exact sequences. The hypothesis that the map between diagrams commutes is essential: if the square involving the $B$-terms fails to commute, a lift used to define $\delta_D$ need not be carried to a lift whose image under the lower vertical map agrees with the construction of $\delta_E$. In that situation the two possible boundary classes can differ in the target cokernel, so there is no induced commutative square for the connecting homomorphisms. Naturality therefore records compatibility of already valid snake diagrams; it does not repair a non-commuting comparison map. ## Kernels and Cokernels of Maps Between Short Exact Sequences A common use of the snake lemma is to compare whether the three vertical maps in a morphism of short exact sequences are injective or surjective. The exact sequence of kernels and cokernels records the precise failure of these properties. [quotetheorem:4533] [citeproof:4533] This form is the one most often quoted in computations. It lets us infer information about the middle vertical map from the left and right vertical maps, and conversely. The inference is valid only because the three horizontal terms form two short exact sequences; without exactness, a kernel in one column need not correspond to a cokernel obstruction in another. For instance, in the failure example above, the element $1\in\mathbb Z$ lies in the right-hand kernel but cannot be lifted through the top map, so there is no boundary class in the left cokernel and no exact kernel-cokernel sequence of the displayed shape. Thus the theorem measures failures of injectivity and surjectivity inside a short exact comparison, not inside an arbitrary commutative diagram. [example: Multiplication by n on the Rational Quotient Sequence] For $n \ge 1$, consider the morphism of short exact sequences of abelian groups obtained by multiplying every term by $n$: \begin{align*} 0 \longrightarrow \mathbb Z \longrightarrow \mathbb Q \longrightarrow \mathbb Q/\mathbb Z \longrightarrow 0 \\ \phantom{0 \longrightarrow{}} \downarrow^{n} \phantom{\mathbb Z} \quad \downarrow^{n} \phantom{\mathbb Q} \quad \downarrow^{n} \phantom{\mathbb Q/\mathbb Z} \\ 0 \longrightarrow \mathbb Z \longrightarrow \mathbb Q \longrightarrow \mathbb Q/\mathbb Z \longrightarrow 0. \end{align*} Multiplication by $n$ is injective on $\mathbb Z$ because \begin{align*} nz=0 &\iff z=0 \end{align*} for $z\in \mathbb Z$, and it is injective on $\mathbb Q$ because \begin{align*} nq=0 &\iff q=0 \end{align*} for $q\in \mathbb Q$. Hence \begin{align*} \ker(n:\mathbb Z\to \mathbb Z)&=0, & \ker(n:\mathbb Q\to \mathbb Q)&=0. \end{align*} Multiplication by $n$ is surjective on $\mathbb Q$ because, for every $q\in\mathbb Q$, \begin{align*} n\left(\frac{q}{n}\right)=q. \end{align*} It is also surjective on $\mathbb Q/\mathbb Z$ because, for every $q+\mathbb Z\in \mathbb Q/\mathbb Z$, \begin{align*} n\left(\frac{q}{n}+\mathbb Z\right) &=q+\mathbb Z. \end{align*} Therefore \begin{align*} \operatorname{coker}(n:\mathbb Q\to \mathbb Q)&=0, & \operatorname{coker}(n:\mathbb Q/\mathbb Z\to \mathbb Q/\mathbb Z)&=0. \end{align*} On the left term, \begin{align*} \operatorname{coker}(n:\mathbb Z\to \mathbb Z) &=\mathbb Z/\operatorname{im}(n:\mathbb Z\to\mathbb Z)\\ &=\mathbb Z/n\mathbb Z. \end{align*} The kernel of multiplication by $n$ on $\mathbb Q/\mathbb Z$ is the $n$-torsion subgroup \begin{align*} (\mathbb Q/\mathbb Z)[n] &=\{q+\mathbb Z\in \mathbb Q/\mathbb Z : n(q+\mathbb Z)=\mathbb Z\}. \end{align*} Thus the sequence supplied by the *Snake Lemma* is \begin{align*} 0 \longrightarrow 0 \longrightarrow 0 \longrightarrow (\mathbb Q/\mathbb Z)[n] \xrightarrow{\delta} \mathbb Z/n\mathbb Z \longrightarrow 0 \longrightarrow 0 \longrightarrow 0. \end{align*} Exactness gives $\ker\delta=0$ and $\operatorname{im}\delta=\mathbb Z/n\mathbb Z$, so $\delta$ is injective and surjective, hence an isomorphism. Now take an element $a/n+\mathbb Z\in(\mathbb Q/\mathbb Z)[n]$ with $a\in\mathbb Z$. It is in the kernel because \begin{align*} n\left(\frac{a}{n}+\mathbb Z\right) &=a+\mathbb Z\\ &=\mathbb Z. \end{align*} To compute $\delta$, lift $a/n+\mathbb Z$ to $a/n\in\mathbb Q$, apply multiplication by $n$, and obtain \begin{align*} n\cdot \frac{a}{n} &=a. \end{align*} Under the inclusion $\mathbb Z\hookrightarrow\mathbb Q$, this element is the integer $a\in\mathbb Z$, so its class in $\operatorname{coker}(n:\mathbb Z\to\mathbb Z)=\mathbb Z/n\mathbb Z$ is \begin{align*} \delta\left(\frac{a}{n}+\mathbb Z\right) &=a+n\mathbb Z. \end{align*} Thus the boundary map identifies the $n$-torsion class represented by $a/n$ with the residue class of the integer obtained after multiplying that rational lift by $n$. [/example] The computation identifies the $n$-torsion of $\mathbb Q/\mathbb Z$ with $\mathbb Z/n\mathbb Z$. It is a compact example of how a quotient term can contain torsion created by a non-surjective map on the submodule. [example: A Commutative Square of Abelian Groups] Consider the square of abelian groups \begin{align*} \mathbb Z \xrightarrow{\times 2} \mathbb Z \\ \downarrow^{\times 3} \quad \downarrow^{\times 3} \\ \mathbb Z \xrightarrow{\times 2} \mathbb Z. \end{align*} It commutes because for every $z\in\mathbb Z$, \begin{align*} (\times 3)((\times 2)(z)) &=(\times 3)(2z)\\ &=6z, \end{align*} and \begin{align*} (\times 2)((\times 3)(z)) &=(\times 2)(3z)\\ &=6z. \end{align*} Each vertical map is injective: if $3z=0$, then $z=0$. Each horizontal map is also injective: if $2z=0$, then $z=0$. For a horizontal map $\times 2:\mathbb Z\to\mathbb Z$, its image is \begin{align*} \operatorname{im}(\times 2) &=\{2z:z\in\mathbb Z\}\\ &=2\mathbb Z, \end{align*} so its cokernel is \begin{align*} \operatorname{coker}(\times 2) &=\mathbb Z/\operatorname{im}(\times 2)\\ &=\mathbb Z/2\mathbb Z. \end{align*} The vertical maps induce a map on the horizontal cokernels \begin{align*} \overline{\times 3}:\mathbb Z/2\mathbb Z&\longrightarrow \mathbb Z/2\mathbb Z, & \overline{\times 3}(z+2\mathbb Z)&=3z+2\mathbb Z. \end{align*} For every $z\in\mathbb Z$, \begin{align*} 3z-z &=2z\in 2\mathbb Z, \end{align*} so \begin{align*} 3z+2\mathbb Z &=z+2\mathbb Z. \end{align*} Thus the induced map on cokernels is the identity map on $\mathbb Z/2\mathbb Z$, hence an isomorphism. By contrast, each vertical map $\times 3:\mathbb Z\to\mathbb Z$ has image \begin{align*} \operatorname{im}(\times 3) &=\{3z:z\in\mathbb Z\}\\ &=3\mathbb Z, \end{align*} and therefore cokernel \begin{align*} \operatorname{coker}(\times 3) &=\mathbb Z/3\mathbb Z. \end{align*} So the induced map on the horizontal cokernels is an isomorphism, even though the vertical maps themselves have nonzero cokernel $\mathbb Z/3\mathbb Z$. This separates the behaviour of a square from the behaviour of its individual arrows: kernels and cokernels must be computed for the specific maps under discussion. [/example] The snake lemma is strongest when embedded into a full short-exact-sequence diagram, but even a single square trains the same habit. First identify which maps determine the kernels and cokernels; then use commutativity to see which induced maps are available. [remark: Diagram Chasing as a Reusable Pattern] A diagram chase is not a substitute for structure; it is a way of using exactness and commutativity without losing track of representatives. The repeated moves are: choose an element in a kernel, lift it across a surjective map, apply a commuting map, replace a zero image by membership in a previous image, and finally check that changing choices changes the answer by an element already being quotiented out. [/remark] The snake lemma closes the gap between short exact sequences and the longer exact sequences that appear later. In Chapter 4, the same connecting-map construction will reappear in homology, where it produces the boundary morphism attached to a short exact sequence of complexes. The snake lemma showed how exactness can produce connecting maps, but it did not yet explain how to compare two large diagrams systematically. The next chapter develops the comparison lemmas, especially the five lemma and its relatives, which let us transfer isomorphism information across exact rows. # 3. The Five Lemma and Related Diagram Lemmas Chapter 2 introduced diagram chasing through the snake lemma: exact rows allow information lost in kernels to reappear in cokernels by a connecting homomorphism. This chapter develops a second family of diagram tools, whose purpose is comparison rather than construction. The guiding question is: when a commutative diagram has exact rows and most vertical maps are known to be injective, surjective, or isomorphisms, what can be forced about the remaining vertical map? The results below are small, but they are used constantly. They turn long verifications into local checks, and they are the mechanism behind many independence-of-choice arguments in homological algebra: construct a map using chosen presentations, resolutions, or liftings, then use a diagram lemma to prove that the result does not depend on the choices. ## Exact Comparison Diagrams A common situation is that two short exact sequences describe the same kind of object, and a morphism between the outer terms has already been understood. The problem is to decide whether the middle map is forced to be an isomorphism. The short five lemma is the basic answer. [explanation: Short Five Lemma] Consider a commutative diagram of $R$-modules with exact rows \begin{align*} 0\longrightarrow A \xrightarrow{i} B \xrightarrow{p} C \longrightarrow 0\\ \phantom{0\longrightarrow{}}\downarrow^{f_A}\phantom{A}\quad \downarrow^{f_B}\phantom{B}\quad \downarrow^{f_C}\phantom{C}\\ 0\longrightarrow \widetilde A \xrightarrow{\widetilde i} \widetilde B \xrightarrow{\widetilde p} \widetilde C \longrightarrow 0. \end{align*} If $f_A$ and $f_C$ are isomorphisms, then $f_B$ is an isomorphism. [/explanation] This argument is the prototype for all diagram chasing in this chapter. The exactness hypotheses convert statements about a middle element into statements about its image on the right and its possible correction by an element from the left. Each hypothesis is doing visible work. If $f_C$ is not injective, an element $b\in B$ may have $p(b)\ne 0$ but $f_C(p(b))=0$, so the first step of the injectivity chase fails. If $f_A$ is not surjective, an error term in $\widetilde{A}$ need not be correctable by something from $A$, so surjectivity of $f_B$ can fail. Exactness is equally essential: without $\ker p=\operatorname{im} i$ and $\ker \widetilde{p}=\operatorname{im}\widetilde{i}$, the proof has no way to identify the obstruction terms as coming from the left-hand modules. Later arguments with homology use exactly this mechanism, with $A$, $B$, and $C$ replaced by cycles, chains, and boundaries. [example: Quotient Module Isomorphism] Let $R$ be a ring, let $M,N$ be left $R$-modules, and let $A\subset M$, $B\subset N$ be submodules. Suppose $f:M\to N$ is an isomorphism and $f(A)=B$. Define \begin{align*} \bar f:M/A&\to N/B,\\ m+A&\mapsto f(m)+B. \end{align*} This formula is well-defined: if $m+A=m'+A$, then $m'-m\in A$, so \begin{align*} f(m')-f(m)=f(m'-m)\in f(A)=B. \end{align*} Hence $f(m')+B=f(m)+B$. The map is $R$-linear because, for $r\in R$ and $m,m'\in M$, \begin{align*} \bar f((m+A)+(m'+A)) &=\bar f(m+m'+A)\\ &=f(m+m')+B\\ &=f(m)+f(m')+B\\ &=(f(m)+B)+(f(m')+B), \end{align*} and \begin{align*} \bar f(r(m+A)) &=\bar f(rm+A)\\ &=f(rm)+B\\ &=r f(m)+B\\ &=r(f(m)+B). \end{align*} The inverse $f^{-1}:N\to M$ satisfies $f^{-1}(B)=A$: if $b\in B$, then $b=f(a)$ for some $a\in A$, so $f^{-1}(b)=a\in A$, and if $a\in A$, then $f(a)\in B$, so $a=f^{-1}(f(a))\in f^{-1}(B)$. Therefore $f^{-1}$ induces \begin{align*} \overline{f^{-1}}:N/B&\to M/A,\\ n+B&\mapsto f^{-1}(n)+A. \end{align*} For every $m\in M$ and $n\in N$, \begin{align*} (\overline{f^{-1}}\circ \bar f)(m+A) &=\overline{f^{-1}}(f(m)+B)\\ &=f^{-1}(f(m))+A\\ &=m+A, \end{align*} and \begin{align*} (\bar f\circ \overline{f^{-1}})(n+B) &=\bar f(f^{-1}(n)+A)\\ &=f(f^{-1}(n))+B\\ &=n+B. \end{align*} Thus $\bar f$ is an isomorphism with inverse $\overline{f^{-1}}$. This example sits next to the short five lemma, but it is not a literal application of that theorem as stated: the map being proved invertible is the right-hand quotient map, while the middle vertical map is already the given isomorphism $f:M\to N$. [/example] The short five lemma gives an isomorphism criterion, but many arguments only know half of the conclusion. The next variants isolate what is needed to prove injectivity or surjectivity of a middle comparison map. [quotetheorem:4534] [citeproof:4534] These variants are useful because comparison maps are often built in stages. At an intermediate stage, injectivity may be available before surjectivity, or surjectivity may be forced by generators before relations have been checked. The asymmetric hypotheses should not be weakened casually. For injectivity, if $f_C$ has a kernel, then an element of $B$ can map to a nonzero element of $C$ that becomes zero after applying $f_C$, so the chase cannot force the element back into $A$. For surjectivity, if $f_A$ is not onto, the final correction term in $\widetilde{A}$ may have no preimage in $A$. These are the same two failure modes as in the short five lemma, separated so that later arguments can use only the half they need. ## The Five Lemma and Four Lemma Long exact sequences introduce a new problem. A map between the middle terms of two exact rows can fail to be controlled by only its immediate neighbours; kernels and cokernels can propagate through several adjacent terms. The five lemma is the standard five-term comparison result. [quotetheorem:1938] [citeproof:1938] This is a longer version of the short five lemma: move right to control obstructions, then move left to correct representatives. The hypotheses split into two groups. Injectivity of $f_3$ uses surjectivity of $f_1$ and injectivity of $f_2,f_4$; surjectivity of $f_3$ uses surjectivity of $f_2,f_4$ and injectivity of $f_5$. Dropping one of these conditions breaks the corresponding chase at a specific point: a missing surjectivity prevents choosing a lift, while a missing injectivity prevents concluding that an obstruction is zero. The surrounding four isomorphisms are therefore stronger than necessary, and the four lemma records the asymmetric hypotheses used by the injective and surjective halves. [quotetheorem:4535] [citeproof:4535] The four lemma is often the form actually used in practice. It tells us which side of the diagram supplies corrections and which side supplies obstructions. The two halves are not interchangeable. In the injective half, $f_1$ must be surjective because the proof needs to pull an element of $B_1$ back to $A_1$ before using injectivity of $f_2$. In the surjective half, $f_5$ must be injective because the proof has to turn $f_5(d_4(a_4))=0$ into $d_4(a_4)=0$. These precise roles explain why long exact homology sequences often require checking different hypotheses on the left and right of the term under comparison. [example: Comparing Presentations] Let $q_{\alpha}:R^n\to M$ and $q_{\beta}:R^t\to M$ be the quotient maps in the two presentations, so \begin{align*} \ker q_{\alpha}&=\operatorname{im}\alpha,\\ \ker q_{\beta}&=\operatorname{im}\beta. \end{align*} Suppose $u:R^n\to R^t$ is $R$-linear and commutes with the quotient maps in the sense that \begin{align*} q_{\beta}u=q_{\alpha}. \end{align*} If $x\in \operatorname{im}\alpha$, then $x=\alpha(a)$ for some $a\in R^m$, and hence \begin{align*} q_{\beta}(u(x)) &=q_{\beta}(u(\alpha(a)))\\ &=q_{\alpha}(\alpha(a))\\ &=0. \end{align*} Since $\ker q_{\beta}=\operatorname{im}\beta$, this gives $u(x)\in \operatorname{im}\beta$. Thus $u$ restricts to a map on relation submodules \begin{align*} \rho:\operatorname{im}\alpha&\to \operatorname{im}\beta,\\ x&\mapsto u(x). \end{align*} The two presentations therefore give a commutative diagram of short exact sequences \begin{align*} 0\to \operatorname{im}\alpha \xrightarrow{i} R^n \xrightarrow{q_{\alpha}} M\to 0,\\ 0\to \operatorname{im}\beta \xrightarrow{j} R^t \xrightarrow{q_{\beta}} M\to 0, \end{align*} where the vertical maps are $\rho$, $u$, and $\operatorname{id}_M$. The left square commutes because, for $x\in \operatorname{im}\alpha$, \begin{align*} (j\rho)(x) &=j(u(x))\\ &=u(x)\\ &=(ui)(x), \end{align*} and the right square commutes because \begin{align*} q_{\beta}u=\operatorname{id}_M q_{\alpha}=q_{\alpha}. \end{align*} If $\rho$ is an isomorphism, then $\rho$ and $\operatorname{id}_M$ are isomorphisms, so the *Short Five Lemma* implies that $u:R^n\to R^t$ is an isomorphism. The same diagram also explains the one-sided versions. Since $\operatorname{id}_M$ is both injective and surjective, the *Short Four Lemma Variants* imply that injectivity of $\rho$ forces injectivity of $u$, while surjectivity of $\rho$ forces surjectivity of $u$. Thus comparison of the relation submodules controls exactly the corresponding comparison of the chosen free generators. [/example] ## Independence of Choices Homological algebra constantly constructs maps by choosing lifts, representatives, splittings, presentations, or resolutions. The next question is not whether a map can be constructed, but whether different choices give canonically equivalent answers. Diagram lemmas provide the verification step: put the two constructions into an exact diagram and prove that the comparison map forced by the universal property is an isomorphism. [example: Induced Map Independent of Representatives] Let $f:M\to N$ be an $R$-linear map with $f(A)\subset B$, where $A\subset M$ and $B\subset N$ are submodules. We define \begin{align*} \bar f:M/A&\to N/B,\\ m+A&\mapsto f(m)+B. \end{align*} To check that this does not depend on the representative $m$, suppose $m+A=m'+A$. Then $m'-m\in A$, so $f(m'-m)\in B$. Since $f$ is $R$-linear, \begin{align*} f(m')-f(m) &=f(m'-m)\in B. \end{align*} Thus $f(m')+B=f(m)+B$, so $\bar f$ is well-defined. The map $\bar f$ is $R$-linear. For $m,m'\in M$, \begin{align*} \bar f((m+A)+(m'+A)) &=\bar f(m+m'+A)\\ &=f(m+m')+B\\ &=f(m)+f(m')+B\\ &=(f(m)+B)+(f(m')+B)\\ &=\bar f(m+A)+\bar f(m'+A), \end{align*} and for $r\in R$, \begin{align*} \bar f(r(m+A)) &=\bar f(rm+A)\\ &=f(rm)+B\\ &=r f(m)+B\\ &=r(f(m)+B)\\ &=r\bar f(m+A). \end{align*} Now suppose $f,g:M\to N$ are $R$-linear maps with $f(A)\subset B$ and $g(A)\subset B$, and suppose also that \begin{align*} (f-g)(M)\subset B. \end{align*} Let $\bar f,\bar g:M/A\to N/B$ be the induced quotient maps. For every $m\in M$, \begin{align*} \bar f(m+A)-\bar g(m+A) &=(f(m)+B)-(g(m)+B)\\ &=(f(m)-g(m))+B\\ &=B, \end{align*} because $f(m)-g(m)\in B$. Hence $\bar f(m+A)=\bar g(m+A)$ for every coset $m+A$, so $\bar f=\bar g$. Thus the quotient map records only the class of $f(m)$ modulo $B$, not the particular representative $m$ nor any part of $f$ whose image already lies in $B$. In larger exact diagrams, this same calculation is the representative-level input that is paired with the *Short Five Lemma* to prove that an induced quotient map is an isomorphism. [/example] The same pattern appears for constructions from presentations. A presentation contains arbitrary choices of generators and relations, but the module it presents should not remember those choices. Diagram lemmas give a precise comparison language for that statement. [remark: Choice Versus Canonicity] A map built from a choice is not automatically non-canonical. The relevant test is whether any two choices are connected by a comparison diagram in which the induced maps on the objects of interest are forced to agree or be isomorphisms. Short exact sequences and five-term exact sequences are the main templates for such comparison diagrams in this course. [/remark] The preceding comparison language still works one short exact sequence at a time. In quotient constructions, however, the data often form a whole grid: rows and columns are exact simultaneously, and exactness must be transferred to the remaining row or column. The obstruction is that a failure of injectivity, surjectivity, or commutativity anywhere in the grid can break the chase, so the full $3\times 3$ configuration needs its own propagation theorem. [quotetheorem:4536] [citeproof:4536] The $3\times 3$ lemma is the local engine behind many arguments that an exact sequence obtained by taking quotients is exact. It explains why diagrams of short exact sequences can be assembled into new short exact sequences. The zeroes at the ends are not decoration. Without injectivity at the left end of the columns or rows, a nonzero element can disappear before the chase begins; without surjectivity at the right end, the initial lift needed for the last term may not exist. Commutativity is also essential: it is what turns information obtained after moving vertically into information about the horizontal maps. This is why the lemma is usually invoked only after a full diagram of short exact sequences has been constructed, rather than from three unrelated exact rows or columns. [illustration:homological-algebra-i-three-by-three-lemma-path] [example: Quotienting A Short Exact Sequence Of Submodules] Let \begin{align*} 0\to A \xrightarrow{i} B \xrightarrow{p} C\to 0 \end{align*} be short exact, and suppose $A_0\subset A$, $B_0\subset B$, and $C_0\subset C$ form a compatible short exact subsequence under the restricted maps: \begin{align*} 0\to A_0 \xrightarrow{i|_{A_0}} B_0 \xrightarrow{p|_{B_0}} C_0\to 0. \end{align*} Define the quotient maps \begin{align*} \overline{i}:A/A_0&\to B/B_0,& a+A_0&\mapsto i(a)+B_0,\\ \overline{p}:B/B_0&\to C/C_0,& b+B_0&\mapsto p(b)+C_0. \end{align*} These are well-defined. If $a+A_0=a'+A_0$, then $a-a'\in A_0$, so \begin{align*} i(a)-i(a')=i(a-a')\in i(A_0)\subset B_0, \end{align*} and hence $i(a)+B_0=i(a')+B_0$. If $b+B_0=b'+B_0$, then $b-b'\in B_0$, so \begin{align*} p(b)-p(b')=p(b-b')\in p(B_0)=C_0, \end{align*} and hence $p(b)+C_0=p(b')+C_0$. We prove that \begin{align*} 0\to A/A_0 \xrightarrow{\overline{i}} B/B_0 \xrightarrow{\overline{p}} C/C_0\to 0 \end{align*} is exact. First, $\overline{i}$ is injective. If $\overline{i}(a+A_0)=B_0$, then $i(a)\in B_0$. Also \begin{align*} p(i(a))=0, \end{align*} so $i(a)\in \ker(p|_{B_0})$. Exactness of the subsequence gives $\ker(p|_{B_0})=i(A_0)$, so there is $a_0\in A_0$ with $i(a)=i(a_0)$. Since $i$ is injective, $a=a_0$, and therefore $a+A_0=A_0$. Next, \begin{align*} (\overline{p}\,\overline{i})(a+A_0) &=\overline{p}(i(a)+B_0)\\ &=p(i(a))+C_0\\ &=0+C_0\\ &=C_0, \end{align*} so $\operatorname{im}\overline{i}\subset \ker\overline{p}$. Conversely, suppose $b+B_0\in \ker\overline{p}$. Then \begin{align*} \overline{p}(b+B_0)=C_0 \end{align*} means $p(b)\in C_0$. Since $p|_{B_0}:B_0\to C_0$ is surjective, choose $b_0\in B_0$ with $p(b_0)=p(b)$. Then \begin{align*} p(b-b_0)=p(b)-p(b_0)=0, \end{align*} so $b-b_0\in \ker p=\operatorname{im}i$. Thus $b-b_0=i(a)$ for some $a\in A$, and \begin{align*} b+B_0 &=(b-b_0)+B_0\\ &=i(a)+B_0\\ &=\overline{i}(a+A_0). \end{align*} Hence $\ker\overline{p}\subset \operatorname{im}\overline{i}$. Finally, $\overline{p}$ is surjective. Given $c+C_0\in C/C_0$, surjectivity of $p:B\to C$ gives $b\in B$ with $p(b)=c$, and then \begin{align*} \overline{p}(b+B_0)=p(b)+C_0=c+C_0. \end{align*} Therefore the quotient row is short exact. The compatibility condition is used exactly in the inclusions $i(A_0)\subset B_0$, $p(B_0)=C_0$, and the identity $\ker(p|_{B_0})=i(A_0)$, which are the points where the quotient maps and the kernel computation depend on the chosen submodules fitting together. [/example] Together, the short five lemma, five lemma, four lemma, and $3\times 3$ lemma form the comparison toolkit for the rest of the course. They will be used when maps of complexes induce maps on homology, when homotopic maps are compared, and when resolutions are shown to be independent of the choices made during their construction. Comparison tools now give a way to control maps between exact sequences of modules and complexes. The next chapter uses them in earnest for short exact sequences of complexes, where the snake lemma produces long exact sequences in homology. # 4. Short Exact Sequences of Complexes This chapter explains how exact sequences interact with homology. In the previous chapters, complexes gave a way to package kernels and images across many degrees, and the snake lemma supplied the diagram chase behind connecting maps. We now combine these ideas: a short exact sequence of complexes gives a long exact sequence on homology, and this long sequence records how cycles in one complex fail to lift to cycles in another. ## Degreewise Exact Sequences of Complexes The first question is what it should mean for several complexes to form an exact sequence. Since a complex is a sequence of modules linked by differentials, exactness has to remember both the module-level exactness in each degree and the compatibility with the differentials. [definition: Short Exact Sequence of Complexes] Let $R$ be a ring, and let $A_*$, $B_*$, and $C_*$ be chain complexes of left $R$-modules. A short exact sequence of chain complexes is a sequence of chain maps \begin{align*} 0 \longrightarrow A_* \xrightarrow{i} B_* \xrightarrow{p} C_* \longrightarrow 0 \end{align*} such that for every $n \in \mathbb Z$ the sequence of $R$-modules \begin{align*} 0 \longrightarrow A_n \xrightarrow{i_n} B_n \xrightarrow{p_n} C_n \longrightarrow 0 \end{align*} is exact. [/definition] The phrase "of complexes" is doing real work: the maps $i$ and $p$ commute with the differentials, so the short exact sequence is not merely a collection of unrelated short exact sequences of modules. [example: Quotient Complex from a Subcomplex] Let $B_*$ be a chain complex and let $A_* \subset B_*$ be a subcomplex, so $A_n \le B_n$ and $d_B(A_n) \subset A_{n-1}$ for every $n$. Define \begin{align*} (B_*/A_*)_n = B_n/A_n. \end{align*} For $b \in B_n$, set \begin{align*} d_{B/A}(b+A_n)=d_B b+A_{n-1}. \end{align*} This is well-defined: if $b+A_n=b'+A_n$, then $b-b' \in A_n$, so \begin{align*} d_B b-d_B b' = d_B(b-b') \in A_{n-1} \end{align*} because $A_*$ is a subcomplex. Hence $d_B b+A_{n-1}=d_B b'+A_{n-1}$. Also, \begin{align*} d_{B/A}^2(b+A_n) &=d_{B/A}(d_B b+A_{n-1})\\ &=d_B^2 b+A_{n-2}\\ &=0+A_{n-2}, \end{align*} so $B_*/A_*$ is a chain complex. The inclusion $i_n:A_n\to B_n$ and quotient map $q_n:B_n\to B_n/A_n$ satisfy \begin{align*} d_B i_n(a)&=d_B a=i_{n-1}(d_A a),\\ d_{B/A}q_n(b)&=d_{B/A}(b+A_n)=d_B b+A_{n-1}=q_{n-1}(d_B b), \end{align*} so they are chain maps. In each degree, \begin{align*} 0 \longrightarrow A_n \xrightarrow{i_n} B_n \xrightarrow{q_n} B_n/A_n \longrightarrow 0 \end{align*} is exact: $i_n$ is injective, $q_n$ is surjective, and \begin{align*} \ker(q_n) &=\{b\in B_n : b+A_n=A_n\}\\ &=\{b\in B_n : b\in A_n\}\\ &=\operatorname{im}(i_n). \end{align*} Therefore the sequence \begin{align*} 0 \longrightarrow A_* \longrightarrow B_* \longrightarrow B_*/A_* \longrightarrow 0 \end{align*} is a short exact sequence of complexes. The subcomplex condition is precisely what lets $d_B$ descend from representatives in $B_n$ to cosets in $B_n/A_n$. [/example] This example is the main model to keep in mind. Even if each degree sequence splits as modules, the splittings need not commute with the differentials, so exactness in the category of complexes is stronger than a degreewise decomposition of modules. [definition: Exact Sequence of Complexes] Let $(X^j_*)_{j \in \mathbb Z}$ be chain complexes and let $f^j: X^j_* \to X^{j+1}_*$ be chain maps. The sequence \begin{align*} \cdots \longrightarrow X^{j-1}_* \xrightarrow{f^{j-1}} X^j_* \xrightarrow{f^j} X^{j+1}_* \longrightarrow \cdots \end{align*} is exact if for every $n \in \mathbb Z$ the degreewise sequence \begin{align*} \cdots \longrightarrow X^{j-1}_n \xrightarrow{f^{j-1}_n} X^j_n \xrightarrow{f^j_n} X^{j+1}_n \longrightarrow \cdots \end{align*} is exact as a sequence of $R$-modules. [/definition] Exactness in complexes is therefore tested degree by degree. Homology, however, is not degreewise: it compares kernels of $d_n$ with images of $d_{n+1}$. The main result of the chapter is that a short exact sequence of complexes produces exactness across homology groups in adjacent degrees. ## Constructing the Connecting Homomorphism The main problem is that a class in $H_n(C_*)$ need not lift to a class in $H_n(B_*)$. It can be represented by a cycle $c \in C_n$, lifted to some $b \in B_n$, but $b$ may fail to be a cycle; the defect $d_B b$ lies in $A_{n-1}$ and defines the boundary of the class. [definition: Connecting Homomorphism] Given a short exact sequence of chain complexes \begin{align*} 0 \longrightarrow A_* \xrightarrow{i} B_* \xrightarrow{p} C_* \longrightarrow 0, \end{align*} the connecting homomorphism in degree $n$ is the map \begin{align*} \partial_n: H_n(C_*) \longrightarrow H_{n-1}(A_*) \end{align*} defined as follows. For a homology class $[c] \in H_n(C_*)$, choose $b \in B_n$ with $p_n(b)=c$. Since $d_C c=0$, exactness gives a unique $a \in A_{n-1}$ satisfying $i_{n-1}(a)=d_B b$. Set \begin{align*} \partial_n([c]) = [a]. \end{align*} [/definition] The formula for $\partial_n$ is only useful if it assembles into a long exact sequence on homology. The possible failure points are exactly the ones suggested by the definition: cycles must map to cycles, boundaries must map to zero at the next stage, and every class killed by one map must come from the previous one. These are the homological exactness questions that the snake-lemma construction resolves degree by degree. [quotetheorem:4537] [citeproof:4537] The sign convention here is the usual chain-complex convention: the connecting map lowers degree by one. For cochain complexes the corresponding connecting map raises degree by one, and the same diagram chase is read with arrows going upward. [example: Boundary in a Quotient Complex] Let $A_* \subset B_*$ be a subcomplex and let $C_* = B_*/A_*$. Write $\bar{b}=b+A_n \in C_n$. If $\bar{b}$ represents a class in $H_n(C_*)$, then it is a cycle in the quotient complex, so \begin{align*} 0 &=d_C(\bar{b})\\ &=d_{B/A}(b+A_n)\\ &=d_B b+A_{n-1}. \end{align*} Thus $d_B b \in A_{n-1}$, because the zero coset in $B_{n-1}/A_{n-1}$ is $A_{n-1}$. In the short exact sequence \begin{align*} 0 \longrightarrow A_* \longrightarrow B_* \longrightarrow B_*/A_* \longrightarrow 0, \end{align*} the quotient map sends $b$ to $\bar{b}$. The connecting homomorphism is defined by lifting the quotient cycle to $b \in B_n$ and then taking the unique element of $A_{n-1}$ whose image in $B_{n-1}$ is $d_B b$. Since $A_* \subset B_*$, this element is exactly $d_B b$, so \begin{align*} \partial_n([\bar{b}])=[d_B b]\in H_{n-1}(A_*). \end{align*} Also $d_B b$ is a cycle in $A_{n-1}$, because \begin{align*} d_A(d_B b)=d_B(d_B b)=d_B^2 b=0. \end{align*} Therefore the boundary map sends a quotient cycle to the actual boundary of any lift in $B_*$, measuring precisely whether the quotient class can be represented by a cycle already lying in $B_n$. [/example] This quotient formula is often the most efficient way to compute connecting maps. It also explains the terminology: the connecting map connects homology of the quotient complex to homology of the subcomplex one degree lower. ## The Long Exact Sequence in Homology The construction of $\partial_n$ suggests a longer pattern. The maps $i$ and $p$ induce homology maps $H_n(i)$ and $H_n(p)$, while the connecting homomorphism moves from $H_n(C_*)$ to $H_{n-1}(A_*)$. The central question is whether these maps fit into an exact sequence. [explanation: Long Exact Homology Sequence For Modules] If \begin{align*} 0\longrightarrow A_* \xrightarrow{i} B_* \xrightarrow{p} C_* \longrightarrow 0 \end{align*} is a short exact sequence of chain complexes of $R$-modules, then there are connecting homomorphisms \begin{align*} \partial_n:H_n(C_*)\longrightarrow H_{n-1}(A_*) \end{align*} such that the sequence \begin{align*} \cdots \longrightarrow H_n(A_*) \xrightarrow{H_n(i)} H_n(B_*) \xrightarrow{H_n(p)} H_n(C_*) \xrightarrow{\partial_n} H_{n-1}(A_*) \longrightarrow \cdots \end{align*} is exact. [/explanation] The long exact sequence is the first major payoff of complexes: a short exact sequence of complexes of modules becomes a global relation among homology groups in all degrees. The connecting homomorphisms carry the information lost by the fact that homology is not an exact functor. [example: Relative Homology Sequence] Suppose $A_* \subset B_*$ is a subcomplex and set $C_* = B_*/A_*$. In each degree the inclusion and quotient maps give an exact sequence \begin{align*} 0 \longrightarrow A_n \longrightarrow B_n \longrightarrow B_n/A_n \longrightarrow 0, \end{align*} because the quotient map is surjective and its kernel is exactly $A_n$. Since $A_*$ is a subcomplex, the quotient differential is \begin{align*} d_{B/A}(b+A_n)=d_Bb+A_{n-1}, \end{align*} so these degreewise short exact sequences assemble into \begin{align*} 0 \longrightarrow A_* \longrightarrow B_* \longrightarrow B_*/A_* \longrightarrow 0. \end{align*} Applying *Long Exact Homology Sequence* to this short exact sequence gives \begin{align*} \cdots \longrightarrow H_n(A_*) \longrightarrow H_n(B_*) \longrightarrow H_n(B_*/A_*) \xrightarrow{\partial_n} H_{n-1}(A_*) \longrightarrow H_{n-1}(B_*) \longrightarrow \cdots . \end{align*} For a quotient cycle $\bar b=b+A_n$ with $d_{B/A}\bar b=0$, we have \begin{align*} 0=d_{B/A}(b+A_n)=d_Bb+A_{n-1}, \end{align*} so $d_Bb\in A_{n-1}$, and the connecting map is \begin{align*} \partial_n([\bar b])=[d_Bb]\in H_{n-1}(A_*). \end{align*} In topological applications, if $B_*$ is the singular chain complex of a space $X$ and $A_*$ is the singular chain complex of a subspace $Y\subset X$, then $H_n(B_*/A_*)$ is written $H_n(X,Y)$. The displayed sequence is the relative homology sequence of the pair $(X,Y)$, and its connecting map records the actual boundary in $Y$ of a relative cycle in $X$ modulo $Y$. [/example] The example also shows why the quotient complex should not be treated as an afterthought. It is the object whose homology records cycles in $B_*$ modulo chains coming from $A_*$, and the connecting map identifies their actual boundary inside $A_*$. ## Naturality of the Long Exact Sequence The next question is whether the long exact sequence is functorial in the short exact sequence of complexes. If we map one short exact sequence of complexes to another, the induced homology maps should commute with the long exact sequences, including the connecting homomorphisms. [definition: Morphism of Short Exact Sequences of Complexes] Let \begin{align*} 0 \longrightarrow A_* \xrightarrow{i} B_* \xrightarrow{p} C_* \longrightarrow 0 \end{align*} and \begin{align*} 0 \longrightarrow A^{\prime}_* \xrightarrow{i^{\prime}} B^{\prime}_* \xrightarrow{p^{\prime}} C^{\prime}_* \longrightarrow 0 \end{align*} be short exact sequences of chain complexes. A morphism from the first sequence to the second is a triple of chain maps $f_A: A_* \to A^{\prime}_*$, $f_B: B_* \to B^{\prime}_*$, and $f_C: C_* \to C^{\prime}_*$ such that \begin{align*} f_B i &= i^{\prime} f_A, & f_C p &= p^{\prime} f_B. \end{align*} [/definition] The two commutative squares say that the maps respect the subcomplexes and quotients. The remaining issue is whether they also respect the boundary information constructed by lifting and differentiating. [quotetheorem:4538] [citeproof:4538] The content of this statement is that the square relating the two connecting homomorphisms commutes: lifting a cycle, applying a comparison map, and differentiating yield the same boundary class regardless of the order in which the steps are carried out. This is not automatic, because the connecting homomorphism is defined through choices of lifts, and one must check that a morphism of short exact sequences carries one system of choices to a compatible one. Once verified, it shows that the connecting map is a genuine natural transformation rather than an artefact of the construction. With the connecting homomorphism shown to be natural, the squares built from $i_*$ and $p_*$ commute for purely functorial reasons, while the square involving the connecting map has just been settled. The natural question is whether these separate compatibilities assemble into a single statement: that a morphism of short exact sequences induces a morphism of the entire long exact sequences, making the two ladders commute rung by rung. [quotetheorem:4539] [citeproof:4539] Naturality of the long exact sequence means it is not merely a sequence attached to an isolated short exact sequence but a construction compatible with all maps between such data. In practice this is what licenses comparison arguments: a map of short exact sequences produces a commuting ladder to which the five lemma and ordinary diagram chases apply, so a complex can be replaced by a more tractable one without losing homological information. In later chapters, this compatibility will allow homotopy equivalent complexes and resolutions to produce compatible maps on derived constructions. ## Mapping Exact Sequences and Computations The final problem is how to use the long exact sequence in practice. Most computations begin with a short exact sequence whose terms are easier to understand than the complex of interest, or with a subcomplex whose quotient isolates the new information. [example: Mapping Exact Sequence for a Subcomplex] Let $A_* \subset B_*$ be a subcomplex. By *Long Exact Homology Sequence* applied to \begin{align*} 0 \longrightarrow A_* \longrightarrow B_* \longrightarrow B_*/A_* \longrightarrow 0, \end{align*} we get the exact segment \begin{align*} H_n(B_*) \longrightarrow H_n(B_*/A_*) \xrightarrow{\partial_n} H_{n-1}(A_*) \longrightarrow H_{n-1}(B_*). \end{align*} Assume $H_n(B_*)=0$ and $H_{n-1}(B_*)=0$. Exactness at $H_n(B_*/A_*)$ gives \begin{align*} \ker(\partial_n) &=\operatorname{im}\bigl(H_n(B_*) \to H_n(B_*/A_*)\bigr)\\ &=\operatorname{im}\bigl(0 \to H_n(B_*/A_*)\bigr)\\ &=0, \end{align*} so $\partial_n$ is injective. Exactness at $H_{n-1}(A_*)$ gives \begin{align*} \operatorname{im}(\partial_n) &=\ker\bigl(H_{n-1}(A_*) \to H_{n-1}(B_*)\bigr)\\ &=\ker\bigl(H_{n-1}(A_*) \to 0\bigr)\\ &=H_{n-1}(A_*), \end{align*} so $\partial_n$ is surjective. Therefore \begin{align*} \partial_n: H_n(B_*/A_*) \longrightarrow H_{n-1}(A_*) \end{align*} is an isomorphism. Thus vanishing of the two neighboring homology groups of $B_*$ shifts the homology of the quotient complex down by one degree into the homology of the subcomplex. [/example] The example illustrates a useful reading habit: vanishing terms in a long exact sequence turn neighbouring maps into injections, surjections, or isomorphisms. The exactness statement is often more valuable than the explicit formula for every map. [remark: Short Exact Sequences Need Not Stay Short on Homology] A short exact sequence of complexes does not usually produce a short exact sequence \begin{align*} 0 \longrightarrow H_n(A_*) \longrightarrow H_n(B_*) \longrightarrow H_n(C_*) \longrightarrow 0. \end{align*} The connecting homomorphism measures the failure of $H_n(p)$ to be surjective and the failure of $H_{n-1}(i)$ to be injective. Thus homology is functorial but not exact in general. [/remark] This failure is not a defect of the theory; it is the phenomenon homological algebra is designed to control. Derived functors later formalize the same idea: when a functor fails to preserve exactness, the surrounding homology groups measure the obstruction. [example: A Two-Term Obstruction] Let $R$ be a ring, and suppose \begin{align*} 0 \longrightarrow A_* \longrightarrow B_* \longrightarrow C_* \longrightarrow 0 \end{align*} is a short exact sequence of complexes with $B_*$ acyclic, so $H_k(B_*)=0$ for every $k$. Applying *Long Exact Homology Sequence* gives the exact segment \begin{align*} H_n(B_*) \longrightarrow H_n(C_*) \xrightarrow{\partial_n} H_{n-1}(A_*) \longrightarrow H_{n-1}(B_*). \end{align*} Since $H_n(B_*)=0$ and $H_{n-1}(B_*)=0$, this becomes \begin{align*} 0 \longrightarrow H_n(C_*) \xrightarrow{\partial_n} H_{n-1}(A_*) \longrightarrow 0. \end{align*} Exactness at $H_n(C_*)$ gives \begin{align*} \ker(\partial_n) &=\operatorname{im}\bigl(H_n(B_*)\to H_n(C_*)\bigr)\\ &=\operatorname{im}\bigl(0\to H_n(C_*)\bigr)\\ &=0, \end{align*} so $\partial_n$ is injective. Exactness at $H_{n-1}(A_*)$ gives \begin{align*} \operatorname{im}(\partial_n) &=\ker\bigl(H_{n-1}(A_*)\to H_{n-1}(B_*)\bigr)\\ &=\ker\bigl(H_{n-1}(A_*)\to 0\bigr)\\ &=H_{n-1}(A_*), \end{align*} so $\partial_n$ is surjective. Hence \begin{align*} \partial_n:H_n(C_*)\longrightarrow H_{n-1}(A_*) \end{align*} is an isomorphism for every $n$. Concretely, if $c\in C_n$ is a cycle, choose $b\in B_n$ mapping to $c$. Since the map $B_*\to C_*$ is a chain map, \begin{align*} p_{n-1}(d_B b) &=d_C(p_n b)\\ &=d_C c\\ &=0. \end{align*} Exactness in degree $n-1$ therefore gives $a\in A_{n-1}$ whose image in $B_{n-1}$ is $d_B b$, and the connecting isomorphism sends \begin{align*} [c]\longmapsto [a]. \end{align*} Thus, when the middle complex is acyclic, every homology class in $C_n$ is measured exactly by the boundary in $A_{n-1}$ of any lift to $B_n$. [/example] This special case appears repeatedly in resolutions. An acyclic complex can still contain meaningful subcomplexes and quotients, and the long exact sequence explains how their homology groups are related by degree shifts. Short exact sequences of complexes show that homology behaves well under extensions, but they also raise a new question: when do two maps of complexes count as the same? The next chapter answers that with chain homotopy, refining equality of induced maps on homology to a stronger notion already visible at the chain level. # 5. Homotopy of Chain Maps This chapter introduces the relation between chain maps that differ by a controlled boundary term. Chapter 1 treated chain maps as the morphisms of complexes and homology as the invariant they induce. Chain homotopy refines this picture: it records when two maps are already equivalent before passing to homology, and it explains why many constructions in homological algebra are insensitive to choices made at the level of complexes. The guiding question is this: when should two maps of complexes be regarded as the same for homological purposes? The answer is not equality of maps degree by degree, but equality up to a family of maps shifting degree by one. The sign convention is part of the structure, because the homotopy must fit the differentials of both complexes. ## Chain Homotopies Between Chain Maps Suppose $C_\bullet$ and $D_\bullet$ are chain complexes of $R$-modules, with differentials $d_C: C_n \to C_{n-1}$ and $d_D: D_n \to D_{n-1}$. If $f,g: C_\bullet \to D_\bullet$ are chain maps, we want a relation saying that $f-g$ is itself built out of differentials. Since $f_n-g_n$ has degree $0$, the correcting maps must have degree $+1$ in the chain-complex convention. [definition: Chain Homotopy] Let $f,g: C_\bullet \to D_\bullet$ be chain maps between chain complexes of $R$-modules. A chain homotopy from $f$ to $g$ is a family of $R$-module homomorphisms \begin{align*} s_n: C_n \to D_{n+1} \end{align*} for all $n \in \mathbb Z$, such that \begin{align*} f_n - g_n = d_D s_n + s_{n-1} d_C \end{align*} as maps $C_n \to D_n$ for every $n$. [/definition] The two terms in the formula have the same total degree as $f_n-g_n$: first move upward by $s_n$ and then downward by $d_D$, or first move downward by $d_C$ and then upward by $s_{n-1}$. This is the chain-complex analogue of saying that a difference is a boundary in the complex of homomorphisms between two complexes. [remark: Sign Convention For Cochain Complexes] For cochain complexes $C^\bullet$ and $D^\bullet$ with differentials of degree $+1$, a cochain homotopy from $f$ to $g$ is usually a family $s^n: C^n \to D^{n-1}$ such that \begin{align*} f^n - g^n = d_D s^n + s^{n+1} d_C. \end{align*} The direction of $s$ changes because cochain differentials raise degree rather than lower degree. [/remark] This convention is worth spelling out because no minus sign appears in the displayed chain formula. The sign has already been absorbed by the choice that chain differentials have degree $-1$ and the homotopy has degree $+1$. [example: Homotopy To Zero In A Short Complex] Let $C_1 \xrightarrow{d} C_0$ and $D_1 \xrightarrow{\partial} D_0$ be chain complexes concentrated in degrees $1$ and $0$, and let $f: C_\bullet \to D_\bullet$ be a chain map. A chain homotopy from $f$ to the zero map is a family $s_n: C_n \to D_{n+1}$ satisfying \begin{align*} f_n=d_Ds_n+s_{n-1}d_C. \end{align*} Only $s_0: C_0 \to D_1$ and $s_1: C_1 \to D_2=0$ can occur in degrees $0$ and $1$. In degree $0$, the differential out of $C_0$ lands in $C_{-1}=0$, so $d_C: C_0 \to C_{-1}$ is the zero map. Hence \begin{align*} f_0 &=d_Ds_0+s_{-1}d_C \\ &=\partial s_0+s_{-1}\cdot 0 \\ &=\partial s_0. \end{align*} In degree $1$, the map $s_1$ has target $D_2=0$, so $d_Ds_1: C_1 \to D_1$ is zero. Therefore \begin{align*} f_1 &=d_Ds_1+s_0d_C \\ &=0+s_0d \\ &=s_0d. \end{align*} Thus null-homotopy in this short complex means that $f_0$ factors through $\partial: D_1 \to D_0$, and once $s_0$ is chosen, the degree $1$ component is forced to be $s_0d$. [/example] The null-homotopy example treats homotopy to the zero map, but later arguments need to compare two arbitrary chain maps in the same way. The relevant relation should remember an explicit family of neighbouring-degree maps whose boundary measures the difference between the two chain maps, rather than merely saying their induced maps on homology agree. [definition: Chain Homotopic Maps] Two chain maps $f,g: C_\bullet \to D_\bullet$ are chain homotopic if there exists a chain homotopy from $f$ to $g$. [/definition] After introducing a relation on maps, the first structural question is whether it behaves like an equality relation. For chain maps this is important because later constructions form categories whose morphisms are homotopy classes of chain maps. [quotetheorem:4540] [citeproof:4540] The equivalence-relation proof uses the additive structure of the Hom groups in an essential way: the inverse homotopy is obtained by negating the family $s$, and transitivity is obtained by adding homotopies. This is why the statement is made for chain maps between fixed complexes of $R$-modules; in a setting without additive inverses, the same formula would not automatically give symmetry. The next issue is compatibility with composition. Without this compatibility, homotopy classes would be only a set of equivalence classes, not morphisms in a usable category. The hypotheses that the maps being composed are chain maps are part of the mechanism, because the proof must move differentials past those maps. [quotetheorem:4541] [citeproof:4541] The theorem does not say that arbitrary degreewise maps preserve homotopies. If $a$ is merely a family of module homomorphisms and not a chain map, then $d_Ca_n$ need not equal $a_{n-1}d_B$, and the precomposition calculation leaves an uncontrolled error term. Likewise, postcomposition needs $b$ to commute with differentials so that $b_nd_Ds_n$ can be rewritten as a boundary in $E_\bullet$. ## Homotopic Maps And Homology The point of chain homotopy is that it kills the ambiguity left by choosing maps on representatives. The problem now is to prove that if $f$ and $g$ differ by a chain homotopy, then cycles receive images that differ by boundaries. This is the basic mechanism behind homotopy invariance in homological algebra. [quotetheorem:1923] [citeproof:1923] This theorem is short, but it is one of the main reasons homotopy is useful. The formula for a homotopy has been engineered so that the second term vanishes on cycles and the first term becomes a boundary. The converse fails: equality on homology does not generally produce a chain homotopy. Let $C_\bullet$ be the exact complex \begin{align*} 0 \to \mathbb Z \xrightarrow{2} \mathbb Z \to \mathbb Z/2\mathbb Z \to 0, \end{align*} with the last nonzero group in degree $-1$. The identity map and the zero map on $C_\bullet$ induce the same maps on homology because all homology groups vanish, but they cannot be chain homotopic; such a homotopy would be a contracting homotopy, and in particular would split the quotient map $\mathbb Z \to \mathbb Z/2\mathbb Z$, which has no $\mathbb Z$-linear section. [example: A Null Homotopic Map Has Zero Induced Map] Let $f: C_\bullet \to D_\bullet$ be chain homotopic to the zero chain map, and let $s_n: C_n \to D_{n+1}$ be a homotopy from $f$ to $0$. Thus, in degree $n$, \begin{align*} f_n-0 &=d_Ds_n+s_{n-1}d_C. \end{align*} If $x\in C_n$ is a cycle, then $d_C(x)=0$, so evaluating the homotopy equation at $x$ gives \begin{align*} f_n(x) &=(f_n-0)(x) \\ &=(d_Ds_n+s_{n-1}d_C)(x) \\ &=d_D(s_n(x))+s_{n-1}(d_C(x)) \\ &=d_D(s_n(x))+s_{n-1}(0) \\ &=d_D(s_n(x)). \end{align*} Since $s_n(x)\in D_{n+1}$, the element $d_D(s_n(x))$ is a boundary in $D_n$. Therefore \begin{align*} H_n(f)([x])=[f_n(x)]=[d_D(s_n(x))]=0 \end{align*} for every homology class $[x]\in H_n(C_\bullet)$. Hence $H_n(f)$ is the zero map in every degree. This is the chain-level reason that a null-homotopic map has no effect on homology. [/example] The converse is not part of the general theory at this level: two maps may induce the same maps on homology without being chain homotopic. Chain homotopy remembers more of the complex than its homology modules, although it forgets enough to preserve all homology groups. ## Homotopy Equivalence Of Complexes Having compared two maps, the next question is when two complexes themselves should be considered the same up to homotopy. Ordinary isomorphism of complexes is often too strict: resolutions, bar constructions, and mapping-cylinder style replacements depend on choices. Homotopy equivalence keeps track of inverse maps only up to chain homotopy. [definition: Homotopy Equivalence] A chain map $f: C_\bullet \to D_\bullet$ is a homotopy equivalence if there exists a chain map $g: D_\bullet \to C_\bullet$ such that \begin{align*} gf &\simeq \operatorname{id}_{C_\bullet}, & fg &\simeq \operatorname{id}_{D_\bullet}, \end{align*} where $\simeq$ denotes chain homotopy. [/definition] The map $g$ is called a homotopy inverse of $f$. The definition asks for inverse behaviour only after passing from chain maps to homotopy classes. [quotetheorem:1925] [citeproof:1925] The theorem gives a hierarchy of notions: isomorphism of complexes implies homotopy equivalence, and homotopy equivalence implies quasi-isomorphism. The converse implication is false without extra hypotheses. For the exact non-contractible complex \begin{align*} 0 \to \mathbb Z \xrightarrow{2} \mathbb Z \to \mathbb Z/2\mathbb Z \to 0, \end{align*} the unique map from this complex to the zero complex is a quasi-isomorphism, since both sides have zero homology, but it is not a homotopy equivalence because that would make the complex contractible. Thus a two-sided homotopy inverse is stronger than merely matching homology groups: it records coherent chain-level inverse data. [example: Deformation Retraction Of A Split Two Term Complex] Let $C_\bullet$ be the two-term complex \begin{align*} C_1=A\oplus B \xrightarrow{d} C_0=B, \qquad d(a,b)=b, \end{align*} and let $H_\bullet$ be the complex with $H_1=A$ and $H_0=0$. Define \begin{align*} i_1(a)&=(a,0), & i_0&=0,\\ p_1(a,b)&=a, & p_0&=0. \end{align*} Then $p i=\operatorname{id}_{H_\bullet}$ degree by degree: in degree $1$, \begin{align*} (p_1 i_1)(a)=p_1(a,0)=a, \end{align*} and in degree $0$ both maps are the zero map $0\to 0$. Now compare $i p$ with $\operatorname{id}_{C_\bullet}$. In degree $1$, \begin{align*} (i_1p_1)(a,b)=i_1(a)=(a,0), \end{align*} while in degree $0$, \begin{align*} (i_0p_0)(b)=0. \end{align*} Define $s_0:C_0=B\to C_1=A\oplus B$ by \begin{align*} s_0(b)=(0,b), \end{align*} and take $s_1=0$ because $C_2=0$. We verify the homotopy equation for $\operatorname{id}_{C_\bullet}-ip$. In degree $0$, \begin{align*} (\operatorname{id}_{C_0}-i_0p_0)(b) &=b-0\\ &=b\\ &=d(0,b)\\ &=(d s_0)(b), \end{align*} and the missing term $s_{-1}d$ is zero because $C_{-1}=0$. In degree $1$, \begin{align*} (\operatorname{id}_{C_1}-i_1p_1)(a,b) &=(a,b)-(a,0)\\ &=(0,b)\\ &=s_0(d(a,b))\\ &=(s_0d)(a,b), \end{align*} and the term $d s_1$ is zero because $s_1=0$. Thus $\operatorname{id}_{C_\bullet}-ip=d s+s d$, so $ip$ is chain homotopic to $\operatorname{id}_{C_\bullet}$. Hence $C_\bullet$ deformation retracts onto the homology complex $H_\bullet$, which keeps the $A$-summand and contracts the $B\xrightarrow{\operatorname{id}}B$ summand. [/example] The splitting example suggests a more structured form of homotopy equivalence. To make that pattern usable later, we need a definition that records not only the existence of a homotopy equivalence but also the direction in which the smaller complex sits inside the larger one. Sometimes a smaller subcomplex is not merely equivalent to the original complex: it sits inside it, admits a projection back onto it, and the only nontrivial comparison is the composite on the larger complex. This is the situation in which a complex can be reduced while retaining an actual inclusion of the reduced model. [definition: Deformation Retraction] A deformation retraction of a chain complex $C_\bullet$ onto a subcomplex $A_\bullet$ consists of chain maps \begin{align*} i: A_\bullet \to C_\bullet, \qquad p: C_\bullet \to A_\bullet \end{align*} such that $pi=\operatorname{id}_{A_\bullet}$ and $ip$ is chain homotopic to $\operatorname{id}_{C_\bullet}$. [/definition] Deformation retractions are stronger than abstract homotopy equivalences because one composite is required to be exactly the identity. In practice this often comes from an explicit splitting of modules degree by degree. ## Contractible Complexes A special case of homotopy equivalence occurs when a complex is homotopy equivalent to the zero complex. The question is what such a condition means internally. It means the identity map itself is a boundary in the Hom complex, so every cycle in the original complex is already a boundary. [definition: Contractible Complex] A chain complex $C_\bullet$ is contractible if the identity chain map $\operatorname{id}_{C_\bullet}: C_\bullet \to C_\bullet$ is chain homotopic to the zero chain map. [/definition] Once contractibility is defined by homotopy of the identity to zero, the immediate question is what this condition forces on homology. A homology class is represented by a cycle, and the contracting identity should express that cycle as a boundary by an explicit neighbouring-degree map. Thus the obstruction to nonzero homology disappears when a contracting homotopy exists. [quotetheorem:4542] [citeproof:4542] The converse is not true for arbitrary complexes of modules. The exact complex \begin{align*} 0 \to \mathbb Z \xrightarrow{2} \mathbb Z \to \mathbb Z/2\mathbb Z \to 0 \end{align*} has zero homology, but it is not contractible: a contracting homotopy would split the surjection $\mathbb Z \to \mathbb Z/2\mathbb Z$, and no such $\mathbb Z$-linear splitting exists. This is the basic limitation: acyclicity says every cycle is some boundary, while contractibility asks for those choices of boundaries to be made by maps compatible in all degrees. [definition: Contracting Homotopy] A contracting homotopy on a chain complex $C_\bullet$ is a family of maps $s_n: C_n \to C_{n+1}$ such that \begin{align*} \operatorname{id}_{C_n}=d_Cs_n+s_{n-1}d_C \end{align*} for every $n \in \mathbb Z$. [/definition] The formula is the chain homotopy equation for a homotopy from the identity to zero. It is useful because it gives a concrete object to construct when proving that a complex is contractible. [example: Augmented Simplex Complex On A Nonempty Set] Let $E$ be a nonempty set and choose $e_\ast \in E$. Form the augmented complex \begin{align*} \cdots \to R[E^{3}] \xrightarrow{d_2} R[E^{2}] \xrightarrow{d_1} R[E] \xrightarrow{\varepsilon} R \to 0, \end{align*} where $R[E^q]$ is the free $R$-module on $q$-tuples of elements of $E$, $\varepsilon([e])=1$, and for $n\geq 1$ the differential is \begin{align*} d_n([e_0,\dots,e_n]) =\sum_{i=0}^{n}(-1)^i[e_0,\dots,\widehat{e_i},\dots,e_n]. \end{align*} Define $s_{-1}:R\to R[E]$ by $s_{-1}(1)=[e_\ast]$, and for $n\geq 0$ define $s_n:R[E^{n+1}]\to R[E^{n+2}]$ on basis elements by \begin{align*} s_n([e_0,\dots,e_n])=[e_\ast,e_0,\dots,e_n]. \end{align*} Since all maps are $R$-linear, it is enough to verify the contracting homotopy equation on basis elements. In degree $-1$, the complex has $C_{-1}=R$, so \begin{align*} (\varepsilon s_{-1})(1) &=\varepsilon([e_\ast])\\ &=1. \end{align*} Thus $\varepsilon s_{-1}=\operatorname{id}_R$. In degree $0$, for a basis element $[e]\in R[E]$, \begin{align*} (d_1s_0+s_{-1}\varepsilon)([e]) &=d_1([e_\ast,e])+s_{-1}(1)\\ &=\bigl([e]-[e_\ast]\bigr)+[e_\ast]\\ &=[e]. \end{align*} Hence $d_1s_0+s_{-1}\varepsilon=\operatorname{id}_{R[E]}$. Now let $n\geq 1$ and take a basis element $[e_0,\dots,e_n]\in R[E^{n+1}]$. First, \begin{align*} d_{n+1}s_n([e_0,\dots,e_n]) &=d_{n+1}([e_\ast,e_0,\dots,e_n])\\ &=[e_0,\dots,e_n] +\sum_{i=0}^{n}(-1)^{i+1}[e_\ast,e_0,\dots,\widehat{e_i},\dots,e_n]. \end{align*} Also, \begin{align*} s_{n-1}d_n([e_0,\dots,e_n]) &=s_{n-1}\left(\sum_{i=0}^{n}(-1)^i[e_0,\dots,\widehat{e_i},\dots,e_n]\right)\\ &=\sum_{i=0}^{n}(-1)^i[e_\ast,e_0,\dots,\widehat{e_i},\dots,e_n]. \end{align*} Adding the two displayed formulas gives \begin{align*} (d_{n+1}s_n+s_{n-1}d_n)([e_0,\dots,e_n]) &=[e_0,\dots,e_n] +\sum_{i=0}^{n}\bigl((-1)^{i+1}+(-1)^i\bigr) [e_\ast,e_0,\dots,\widehat{e_i},\dots,e_n]\\ &=[e_0,\dots,e_n], \end{align*} because $(-1)^{i+1}+(-1)^i=0$ for every $i$. Therefore $d s+s d=\operatorname{id}$ in every degree, so the augmented simplex complex is contractible. [/example] The chosen basis element is not canonical, and that is the point of the example. A contraction can depend on choices even when the conclusion about homology is canonical. [example: Split Exact Complex Is Contractible In Two Terms] Let $0 \to A \xrightarrow{i} B \xrightarrow{p} C \to 0$ be split exact, and choose $t:C\to B$ with $pt=\operatorname{id}_C$. First suppose $A=0$, so $\ker(p)=\operatorname{im}(i)=0$. Thus $p$ is injective. For any $b\in B$, \begin{align*} p(tp(b))&=(pt)(p(b))\\ &=\operatorname{id}_C(p(b))\\ &=p(b), \end{align*} so injectivity of $p$ gives $tp(b)=b$. Hence $tp=\operatorname{id}_B$. View the complex as $C_1=B$, $C_0=C$, and $d=p$. Define \begin{align*} s_0&=t:C\to B, & s_1&=0. \end{align*} In degree $0$, the contracting homotopy equation is \begin{align*} (d s_0+s_{-1}d)(c) &=p(t(c))+0\\ &=(pt)(c)\\ &=c. \end{align*} In degree $1$, since $C_2=0$, the term $d s_1$ is zero, and \begin{align*} (d s_1+s_0d)(b) &=0+t(p(b))\\ &=(tp)(b)\\ &=b. \end{align*} Thus $d s+s d=\operatorname{id}$, so the two-term complex $B\xrightarrow{p}C$ is contractible when $A=0$. For the general split short exact sequence, every $b\in B$ decomposes as \begin{align*} b &=\bigl(b-tp(b)\bigr)+tp(b), \end{align*} and \begin{align*} p\bigl(b-tp(b)\bigr) &=p(b)-ptp(b)\\ &=p(b)-p(b)\\ &=0. \end{align*} Thus $b-tp(b)\in \ker(p)=\operatorname{im}(i)$, while $tp(b)$ lies in the chosen copy of $C$ inside $B$. Under this splitting, the complex $B\xrightarrow{p}C$ is the direct sum of \begin{align*} A\to 0 \qquad\text{and}\qquad C\xrightarrow{\operatorname{id}_C}C. \end{align*} The second summand contracts by the same formula $s_0=\operatorname{id}_C$ and $s_1=0$, while the first summand remains as the homology in degree $1$. This is the local algebra behind deformation retractions of split complexes. [/example] Contractibility is therefore the homotopical version of having no homology together with explicit choices of primitives. In later chapters this language will make resolutions manageable: different projective resolutions of the same module will be compared by maps unique up to chain homotopy, so their derived constructions will not depend on the chosen resolution. Chain homotopy tells us when two chain maps differ by a controlled boundary term, but it still does not measure how far a map is from being a homology equivalence. The next chapter packages that defect into mapping cones and uses them to define quasi-isomorphisms. # 6. Mapping Cones and Quasi-Isomorphisms Mapping cones package the failure of a chain map to be a homology equivalence into a single complex. Earlier chapters used short exact sequences and homotopies to compare complexes; this chapter adds a construction that turns a map $f:C \to D$ into a new complex whose homology measures exactly what $f$ misses. The key result is the cone exact sequence, from which the standard test follows: $f$ is a quasi-isomorphism precisely when $\operatorname{Cone}(f)$ is acyclic. ## The Mapping Cone of a Chain Map Given a chain map $f:C \to D$, the first problem is to build a complex that remembers both the target $D$ and a shifted copy of the source $C$. The shift is necessary because a class of $C$ should appear as the next boundary data for a class of $D$, and the sign is chosen so that the resulting differential squares to zero. [definition: Mapping Cone] Let $C_\bullet$ and $D_\bullet$ be chain complexes of $R$-modules, with differentials $d_C$ and $d_D$, and let $f:C_\bullet \to D_\bullet$ be a chain map. The mapping cone of $f$ is the chain complex $\operatorname{Cone}(f)$ defined by \begin{align*} \operatorname{Cone}(f)_n &= D_n \oplus C_{n-1}, \\ d_{\operatorname{Cone}(f)}(y,x) &= (d_D y + f(x), -d_C x) \end{align*} for $y \in D_n$ and $x \in C_{n-1}$. [/definition] The cone is meant to be a chain complex, so the defining formula still has to pass the basic test $d^2=0$. The only possible obstruction comes from the mixed terms involving the chain map $f$: after applying the differential twice, the two contributions of $f(d_Cx)$ must cancel. The sign in the shifted source component is chosen exactly to remove that obstruction. [quotetheorem:4543] [citeproof:4543] This construction is functorial enough to support diagram arguments, but its first use is computational. The cone of the identity map should have no homology, because the identity map loses no information. [example: Cone of the Identity] Let $C$ be any chain complex and consider $\operatorname{id}_C:C \to C$. By the definition of the mapping cone, \begin{align*} \operatorname{Cone}(\operatorname{id}_C)_n &= C_n\oplus C_{n-1}, \\ d_n(y,x) &=(d_Cy+x,-d_Cx) \end{align*} for $y\in C_n$ and $x\in C_{n-1}$. We show that this cone is contractible by defining \begin{align*} s_n:\operatorname{Cone}(\operatorname{id}_C)_n &\longrightarrow \operatorname{Cone}(\operatorname{id}_C)_{n+1}, \\ s_n(y,x)&=(0,y). \end{align*} Now compute the two terms in the chain homotopy identity. First, \begin{align*} d_{n+1}s_n(y,x) &=d_{n+1}(0,y) \\ &=(d_C0+y,-d_Cy) \\ &=(y,-d_Cy). \end{align*} Second, \begin{align*} s_{n-1}d_n(y,x) &=s_{n-1}(d_Cy+x,-d_Cx) \\ &=(0,d_Cy+x). \end{align*} Adding these two expressions gives \begin{align*} (d_{n+1}s_n+s_{n-1}d_n)(y,x) &=(y,-d_Cy)+(0,d_Cy+x) \\ &=(y,x). \end{align*} Thus $ds+sd=\operatorname{id}_{\operatorname{Cone}(\operatorname{id}_C)}$, so the identity map on the cone is chain homotopic to the zero map. Therefore the cone of the identity map is stronger than acyclic: it is contractible, hence chain homotopy equivalent to the zero complex. [/example] The next example shows that cones also recover familiar quotient modules from very small complexes. It is the prototype for viewing cokernels as homology groups. [example: Cone of Multiplication by n] Let $n\ge 1$ and let $f:\mathbb Z[0]\to \mathbb Z[0]$ be multiplication by $n$, where $\mathbb Z[0]_0=\mathbb Z$ and $\mathbb Z[0]_k=0$ for $k\ne 0$. By the definition of the mapping cone, \begin{align*} \operatorname{Cone}(f)_m &=\mathbb Z[0]_m\oplus \mathbb Z[0]_{m-1}. \end{align*} Thus \begin{align*} \operatorname{Cone}(f)_1&=0\oplus \mathbb Z\cong \mathbb Z,\\ \operatorname{Cone}(f)_0&=\mathbb Z\oplus 0\cong \mathbb Z, \end{align*} and all other degrees are zero. Under these identifications, an element of $\operatorname{Cone}(f)_1$ has the form $(0,x)$ with $x\in\mathbb Z$. Its boundary is \begin{align*} d_1(0,x) &=(d_{\mathbb Z[0]}0+f(x),-d_{\mathbb Z[0]}x)\\ &=(0+nx,0)\\ &=(nx,0). \end{align*} So the cone is the two-term complex \begin{align*} 0 \longrightarrow \mathbb Z \xrightarrow{n} \mathbb Z \longrightarrow 0, \end{align*} with the left copy in degree $1$ and the right copy in degree $0$. Now compute its homology. Since $d_2:0\to \mathbb Z$ is the zero map and $d_1:\mathbb Z\to\mathbb Z$ is multiplication by $n$, \begin{align*} H_1(\operatorname{Cone}(f)) &=\ker(d_1)/\operatorname{im}(d_2)\\ &=\{x\in\mathbb Z:nx=0\}/0\\ &=0, \end{align*} because $n\ge 1$ and $\mathbb Z$ has no nonzero element killed by $n$. Also $d_0:\mathbb Z\to 0$ is zero, so \begin{align*} H_0(\operatorname{Cone}(f)) &=\ker(d_0)/\operatorname{im}(d_1)\\ &=\mathbb Z/n\mathbb Z. \end{align*} The cone has converted the map $n:\mathbb Z\to\mathbb Z$ into the quotient module it presents. [/example] ## The Cone Exact Sequence The cone contains $D$ as the first summand and a shifted copy of $C$ as the second summand. The natural question is whether this decomposition gives a short exact sequence of complexes, because short exact sequences are precisely the input for the connecting homomorphism machinery developed earlier. [explanation: Mapping Cone Short Exact Sequence] With the shift convention $C[k]_n=C_{n-k}$, the mapping cone fits into a short exact sequence of chain complexes \begin{align*} 0\longrightarrow D \longrightarrow \operatorname{Cone}(f) \longrightarrow C[1]\longrightarrow 0. \end{align*} The first map sends $y\in D_n$ to $(y,0)\in D_n\oplus C_{n-1}$, and the second map sends $(y,x)$ to $x\in C_{n-1}=C[1]_n$. [/explanation] This short exact sequence is the structural payoff of the cone construction: it packages the chain map $f$ together with its target into a single exact sequence of complexes, exactly the form required to feed the connecting-homomorphism machinery. Because the sequence is built from the mapping cone rather than from an arbitrary choice of subcomplex and quotient, the data it records is intrinsic to $f$ rather than to incidental choices, which is what makes the resulting invariants meaningful. The short exact sequence becomes useful only once its long exact homology sequence is written out, and the decisive point is to identify what the connecting map computes. Since that connecting map recovers the map induced by $f$ on homology, the long exact sequence relates $H_*(C)$, $H_*(D)$, and $H_*(\operatorname{Cone}(f))$ directly through $f_*$. The following exact sequence is the point where the cone stops being only a construction and becomes a diagnostic device for the homological behaviour of $f$. [explanation: Cone Long Exact Sequence] The short exact sequence \begin{align*} 0\longrightarrow D \longrightarrow \operatorname{Cone}(f) \longrightarrow C[1]\longrightarrow 0 \end{align*} induces the long exact sequence \begin{align*} \cdots \longrightarrow H_n(C) \xrightarrow{H_n(f)} H_n(D) \longrightarrow H_n(\operatorname{Cone}(f)) \longrightarrow H_{n-1}(C) \xrightarrow{H_{n-1}(f)} H_{n-1}(D) \longrightarrow \cdots . \end{align*} [/explanation] The sequence says that the homology of the cone measures the failure of $f$ to be an isomorphism on homology: it sits in an exact sequence between the homology of $C$ and that of $D$, with the comparison maps governed by $f_*$. When the cone homology vanishes in every degree, exactness forces each induced map $f_*$ to be an isomorphism, which is precisely the criterion exploited in the next section to detect quasi-isomorphisms. Conversely, any nonzero cone homology records a degree in which the comparison has a kernel or cokernel defect, so the cone is not merely auxiliary bookkeeping. ## Quasi-Isomorphisms Detected by Cones A quasi-isomorphism is a chain map that becomes an isomorphism after passing to homology. The cone exact sequence gives a practical way to test this condition without comparing kernels and cokernels degree by degree. [definition: Quasi-Isomorphism] Let $f:C\to D$ be a chain map of chain complexes. The map $f$ is a quasi-isomorphism if, for every $n\in\mathbb Z$, the induced map \begin{align*} H_n(f):H_n(C)\longrightarrow H_n(D) \end{align*} is an isomorphism. [/definition] The cone construction was introduced to replace the many maps $H_n(f)$ by one object whose homology records all their defects at once. The theorem below is the payoff: it translates the quasi-isomorphism question for $f$ into the acyclicity of a single complex. [explanation: Cone Criterion For Quasi-Isomorphisms] For a chain map $f:C\to D$ of chain complexes of $R$-modules, the map $f$ is a quasi-isomorphism if and only if the mapping cone $\operatorname{Cone}(f)$ is acyclic, that is, \begin{align*} H_n(\operatorname{Cone}(f))=0 \end{align*} for every $n$. [/explanation] This criterion is often the first reason mapping cones appear in homological algebra. Instead of saying that $C$ and $D$ have the same homology through $f$, we say that the defect complex $\operatorname{Cone}(f)$ has no homology. [remark: Acyclic Versus Contractible] The cone criterion only requires acyclicity, not contractibility. The cone of the identity map is contractible, but the cone of a quasi-isomorphism need not be contractible unless stronger homotopy information is available. This distinction matters because quasi-isomorphism is weaker than chain homotopy equivalence. [/remark] The criterion also explains the multiplication example. The map $n:\mathbb Z[0]\to\mathbb Z[0]$ is not a quasi-isomorphism for $n>1$, and the cone records the obstruction as $H_0\cong\mathbb Z/n\mathbb Z$. ## Mapping Cylinders and Homotopy Intuition The mapping cone measures the failure of $f$ to be a quasi-isomorphism. A related construction, the mapping cylinder, keeps the map $f:C\to D$ visible as a factorisation through a larger complex, mirroring the topological mapping cylinder without leaving the category of chain complexes. [definition: Mapping Cylinder] Let $f:C\to D$ be a chain map. A mapping cylinder for $f$ is the chain complex \begin{align*} \operatorname{Cyl}(f)_n=D_n\oplus C_n\oplus C_{n-1} \end{align*} with differential \begin{align*} d_{\operatorname{Cyl}(f)}(y,x,s)=(d_Dy-f(s),d_Cx+s,-d_Cs). \end{align*} [/definition] The cylinder is useful only if these informal endpoint maps actually interact correctly with the differential. There are three compatibility checks to make: the two endpoint inclusions must be chain maps, the projection to $D$ must be a chain map, and the projection should collapse the cylinder in a way that recovers $f$ on the $C$ endpoint. These checks turn the displayed complex into a usable factorization device. [quotetheorem:4547] [citeproof:4547] The cylinder viewpoint separates two ideas that the cone compresses together: adjoining a formal homotopy direction and then collapsing it. This is why cylinders are useful for intuition even when the cone is the main computational object. [example: Cylinder of the Identity] For $f=\operatorname{id}_C$, the mapping cylinder has \begin{align*} \operatorname{Cyl}(\operatorname{id}_C)_n &=C_n\oplus C_n\oplus C_{n-1},\\ d(y,x,s) &=(d_Cy-s,d_Cx+s,-d_Cs), \end{align*} because $f(s)=s$. The two endpoint inclusions are \begin{align*} j_D(y)&=(y,0,0),\\ j_C(x)&=(0,x,0), \end{align*} and the projection is \begin{align*} q(y,x,s)=y+x. \end{align*} Applying $q$ to the two endpoint copies gives \begin{align*} qj_D(y)&=q(y,0,0)=y+0=y,\\ qj_C(x)&=q(0,x,0)=0+x=x. \end{align*} Thus both endpoint copies map back to the same complex $C$ by the identity map. The third summand is the algebraic homotopy direction because its contribution cancels under $q$. Indeed, \begin{align*} q(d(y,x,s)) &=q(d_Cy-s,d_Cx+s,-d_Cs)\\ &=(d_Cy-s)+(d_Cx+s)\\ &=d_Cy+d_Cx\\ &=d_C(y+x)\\ &=d_C(q(y,x,s)). \end{align*} So the cylinder of the identity contains two copies of $C$ connected by a shifted copy of $C$, and projection collapses this interval object back onto $C$ while identifying both endpoints with the identity. [/example] Mapping cones are often drawn from cylinders by collapsing the target copy of $D$. At the level of complexes, the exact sequence for the cone is the algebraic remnant of that collapse: the source has shifted degree, the target remains in place, and the boundary remembers $f$. Mapping cones turn homological failure into a single auxiliary complex, which is exactly the kind of construction needed for resolution theory. The next chapter introduces projective modules, the objects that make lifting arguments possible when we begin building resolutions. # 7. Projective Modules Projective modules enter the course at the point where exact sequences and diagram lemmas become construction tools. Earlier chapters used free modules as convenient sources of generators; this chapter isolates the property of free modules that matters for lifting maps across surjections. The central theme is that projective modules are precisely the modules for which maps out of them behave as if quotients had no obstruction. ## Lifting Along Surjections When a map $M \twoheadrightarrow N$ forgets information, a map into $N$ need not remember enough data to come from a map into $M$. Projectivity is the condition that a chosen source module never creates such an obstruction. [definition: Projective Module] Let $R$ be a ring. A left $R$-module $P$ is projective if for every surjective $R$-module homomorphism $q: M \to N$ and every $R$-module homomorphism $f: P \to N$, there exists an $R$-module homomorphism $\tilde f: P \to M$ such that $q \circ \tilde f = f$. [/definition] The map $\tilde f$ is called a lift of $f$ along $q$. The definition says existence of lifts, not uniqueness: if $\ker q$ is non-zero, different lifts may differ by a map $P \to \ker q$. [example: Free Modules Lift Across Surjections] Let $F=\bigoplus_{i\in I}Re_i$ be a free left $R$-module, let $q:M\to N$ be surjective, and let $f:F\to N$ be $R$-linear. For each $i\in I$, surjectivity of $q$ gives an element $m_i\in M$ such that $q(m_i)=f(e_i)$. Since $F$ is free on the basis $\{e_i\}_{i\in I}$, there is a unique $R$-linear map $\tilde f:F\to M$ satisfying $\tilde f(e_i)=m_i$ for every $i$. To verify that $\tilde f$ is a lift of $f$, take an arbitrary element $x\in F$. Because $F$ is a direct sum, $x$ has a finite expression \begin{align*} x=\sum_{j=1}^r a_j e_{i_j} \end{align*} with $a_j\in R$. Using $R$-linearity of $\tilde f$, $q$, and $f$, we compute \begin{align*} (q\circ \tilde f)(x) &=q\left(\tilde f\left(\sum_{j=1}^r a_j e_{i_j}\right)\right)\\ &=q\left(\sum_{j=1}^r a_j \tilde f(e_{i_j})\right)\\ &=q\left(\sum_{j=1}^r a_j m_{i_j}\right)\\ &=\sum_{j=1}^r a_j q(m_{i_j})\\ &=\sum_{j=1}^r a_j f(e_{i_j})\\ &=f\left(\sum_{j=1}^r a_j e_{i_j}\right)\\ &=f(x). \end{align*} Thus $q\circ \tilde f=f$, so every map from a free module lifts across every surjective module homomorphism. [/example] The example gives the main supply of projective modules: every free module lifts maps across surjections. This raises the question of whether the lifting property pins down a recognisable class of modules or is too weak to characterise anything concrete. The natural guess is that the modules with this property are exactly the direct summands of free modules, since a summand inherits liftings from the ambient free module by composing with the projection onto it. The following theorem confirms this and adds a third, functorial description in terms of $\operatorname{Hom}_R$, so that the same class can be detected in whichever language a given problem makes most convenient. [quotetheorem:4548] [citeproof:4548] This theorem is the main working dictionary, and its value lies in the three descriptions being interchangeable while differing in strength. In computations, direct summands of free modules are often easier to recognise than the lifting property, while in proofs about exactness the $\operatorname{Hom}$ formulation is the natural language. Having several equivalent definitions also makes projectivity robust under the operations of homological algebra: a property checkable through summands, liftings, or exactness of $\operatorname{Hom}$ tends to be preserved by the constructions that respect any one of those formulations. ## Direct Summands And Idempotents A direct summand of a free module is not only an abstract submodule; for finite free modules it can be encoded by an idempotent matrix. This gives an algebraic way to see projective modules that are not initially presented by generators and relations. [definition: Idempotent Endomorphism] Let $M$ be a left $R$-module. An endomorphism $e: M \to M$ is idempotent if $e^2=e$. [/definition] An idempotent endomorphism behaves like a projection, but this analogy has to be converted into an actual module decomposition. The key question is whether every element of $M$ can be separated uniquely into a part fixed by $e$ and a part killed by $e$. If so, idempotents give a concrete way to manufacture direct summands and hence projective modules. [quotetheorem:4549] [citeproof:4549] For $M=R^n$, endomorphisms are represented by $n\times n$ matrices with entries in the opposite ring convention if modules are written as columns over a noncommutative ring. In the commutative case this distinction disappears, and idempotent matrices give concrete projective summands of $R^n$. [example: Idempotent Matrix Over A Product Ring] Let $R=k\times k$ and let $e=(1,0)$. Multiplication in $R$ is componentwise, so \begin{align*} e^2 &=(1,0)(1,0)\\ &=(1\cdot 1,0\cdot 0)\\ &=(1,0)\\ &=e. \end{align*} Thus multiplication by $e$ defines an idempotent endomorphism $R\to R$. For an element $(a,b)\in R$, the image of this endomorphism is computed by \begin{align*} e(a,b) &=(1,0)(a,b)\\ &=(a,0). \end{align*} Hence \begin{align*} \operatorname{im}(e) =\{(a,0):a\in k\} =k\times 0 =eR. \end{align*} The kernel consists of those pairs killed by multiplication by $e$: \begin{align*} e(a,b)=0 &\Longleftrightarrow (a,0)=(0,0)\\ &\Longleftrightarrow a=0. \end{align*} Therefore \begin{align*} \ker(e)=\{(0,b):b\in k\}=0\times k=(1-e)R, \end{align*} since $1-e=(1,1)-(1,0)=(0,1)$ and $(0,1)(a,b)=(0,b)$. Every element of $R$ decomposes as \begin{align*} (a,b)=(a,0)+(0,b), \end{align*} and the intersection is zero because $(a,0)=(0,b)$ forces $a=0$ and $b=0$. Thus \begin{align*} R=eR\oplus (1-e)R. \end{align*} So $eR\cong k\times 0$ is projective as an $R$-module because it is a direct summand of the free module $R$. It is not generated by a unit of $R$: the units in $k\times k$ are exactly pairs $(u,v)$ with $u\ne 0$ and $v\ne 0$, while every element of $eR$ has second coordinate $0$. [/example] This example shows why projective modules are best understood through splitting rather than through bases. A projective module may be locally or piecewise free without possessing a global basis. ## Examples Over Principal Ideal Domains And Local Rings The next question is how far projective modules are from free modules in familiar algebraic settings. Over some rings the distinction disappears for finitely generated modules; over other rings idempotents and local data create genuine new examples. [quotetheorem:4550] [citeproof:4550] The theorem explains why examples over $\mathbb Z$ may hide the difference between free and projective in the finitely generated torsion-free case. Torsion modules give the opposite behaviour. [example: The Module Z Mod N Is Not Projective] Let $n\ge 2$, and let $q:\mathbb Z\to \mathbb Z/n\mathbb Z$ be the quotient homomorphism. We show that $\mathbb Z/n\mathbb Z$ is not projective as a $\mathbb Z$-module. Suppose, for contradiction, that $\mathbb Z/n\mathbb Z$ is projective. Applying the lifting property to the surjection $q$ and the identity map $\operatorname{id}_{\mathbb Z/n\mathbb Z}$ gives a $\mathbb Z$-linear map \begin{align*} s:\mathbb Z/n\mathbb Z\to \mathbb Z \end{align*} such that \begin{align*} q\circ s=\operatorname{id}_{\mathbb Z/n\mathbb Z}. \end{align*} Let \begin{align*} a=s(\bar 1)\in \mathbb Z. \end{align*} Then \begin{align*} \bar a =q(a) =q(s(\bar 1)) =(q\circ s)(\bar 1) =\operatorname{id}_{\mathbb Z/n\mathbb Z}(\bar 1) =\bar 1, \end{align*} so $a\equiv 1\pmod n$. On the other hand, $\bar n=\bar 0$ in $\mathbb Z/n\mathbb Z$, so \begin{align*} 0 &=s(\bar 0)\\ &=s(\bar n)\\ &=s(n\bar 1)\\ &=n\,s(\bar 1)\\ &=na, \end{align*} where the fourth equality uses $\mathbb Z$-linearity of $s$. Since $\mathbb Z$ is an integral domain and $n\ge 2$, the equality $na=0$ forces $a=0$. But $a=0$ gives $a\equiv 0\pmod n$, contradicting $a\equiv 1\pmod n$. Therefore $\mathbb Z/n\mathbb Z$ is not projective. [/example] Local rings give a different test: finite projective modules over them have bases. This result is useful because many geometric and commutative algebra arguments reduce projectivity to a local freeness condition. [quotetheorem:4551] [citeproof:4551] Together with the result that finitely generated projective modules over a PID are free, the local-ring theorem shows how strongly the base ring controls the gap between projective and free. Over both PIDs and local rings the two notions coincide, which is exactly why the distinction stays invisible in the first examples one meets in algebra. The lesson is that freeness is the wrong invariant to carry forward: it is an accident of especially well-behaved rings, whereas splitting is the property that survives base change, localisation, and exactness of $\operatorname{Hom}$, and is therefore the notion that generalises to the rings where projective and free genuinely differ. ## Projective Covers And Free Presentations Free modules always provide surjections, but they may be much larger than necessary. A projective cover is a minimal projective module mapping onto a given module, when such a minimal object exists. [definition: Superfluous Submodule] Let $M$ be a left $R$-module. A submodule $K\subset M$ is superfluous if for every submodule $L\subset M$, the equality $K+L=M$ implies $L=M$. [/definition] Superfluous kernels formalise the idea that no non-zero part of the source can be discarded while retaining surjectivity. This leads to the version of minimality used for projective covers. [definition: Projective Cover] Let $R$ be a ring and let $M$ be a left $R$-module. A projective cover of $M$ is a surjective homomorphism $p:P\to M$ such that $P$ is projective and $\ker p$ is superfluous in $P$. [/definition] Projective covers are powerful when available, but they are not part of the basic existence theory over arbitrary rings. The next issue is therefore existence rather than minimality: before constructing resolutions, we need a dependable source of surjections onto arbitrary modules. The construction that always exists is the free presentation, and it begins with the basic fact that every module can be reached from a free module. [quotetheorem:4552] [citeproof:4552] This theorem is the reason free modules suffice for many constructions in the first pass through homological algebra: every module is a quotient of a free one, so a surjection from a free module is always available even when no projective cover exists. The trade-off is that this surjection is far from unique and usually far from minimal, so the resulting presentations carry redundant generators and relations. For derived-functor constructions that redundancy is harmless, because the later arguments require exactness rather than minimality, and a free module can always start a presentation or resolution. [definition: Free Presentation] Let $R$ be a ring and let $M$ be a left $R$-module. A free presentation of $M$ is an exact sequence \begin{align*} F_1 \longrightarrow F_0 \longrightarrow M \longrightarrow 0 \end{align*} where $F_0$ and $F_1$ are free left $R$-modules. [/definition] The first free module records generators and the second records relations. Iterating this idea produces free resolutions, while replacing free modules by arbitrary projective modules produces projective resolutions. [remark: Covers Versus Presentations] Over semiperfect rings, finitely generated modules have projective covers, and these covers are often the right minimal objects. Over a general ring, projective covers may fail to exist, but free presentations still exist for every module. For derived functor constructions, existence and exactness matter more than minimality, so free resolutions usually suffice at this stage of the course. [/remark] ## Exactness Of Hom From A Projective Module The final point of the chapter connects projectivity back to exact sequences. Since $\operatorname{Hom}_R(P,-)$ converts module maps out of $P$ into abelian groups of maps, the question is exactly how much exactness survives. [explanation: Left Exactness Of Hom From A Module] For any left $R$-module $P$, applying $\operatorname{Hom}_R(P,-)$ to a short exact sequence \begin{align*} 0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0 \end{align*} gives an exact sequence of abelian groups \begin{align*} 0\longrightarrow \operatorname{Hom}_R(P,A) \longrightarrow \operatorname{Hom}_R(P,B) \longrightarrow \operatorname{Hom}_R(P,C). \end{align*} The final map need not be surjective. [/explanation] The content of this statement is that $\operatorname{Hom}_R(P,-)$ never destroys injectivity, nor exactness in the middle of a short exact sequence; what can fail is surjectivity at the right-hand end, where a map out of $P$ into a quotient need not lift to the total module. This left exactness holds for $P$ in the first slot regardless of any special property of $P$, so it isolates the single place where the functor is able to lose information. Because left exactness is automatic, any failure of $\operatorname{Hom}_R(P,-)$ to be exact is concentrated entirely in whether surjections are preserved. That is precisely the lifting condition built into projectivity. The theorem below turns this observation into a characterisation: projective modules are exactly those $P$ for which the Hom functor preserves the missing right-hand exactness and hence becomes fully exact. [explanation: Projective Modules And Exactness Of Hom] A left $R$-module $P$ is projective if and only if $\operatorname{Hom}_R(P,-)$ preserves surjections. Equivalently, for every short exact sequence of left $R$-modules \begin{align*} 0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0, \end{align*} the induced sequence \begin{align*} 0\longrightarrow \operatorname{Hom}_R(P,A) \longrightarrow \operatorname{Hom}_R(P,B) \longrightarrow \operatorname{Hom}_R(P,C) \longrightarrow 0 \end{align*} is exact. [/explanation] This result is the bridge to resolutions. A projective resolution is useful because projective terms preserve the surjections needed to control exactness, while the earlier existence theorem ensures that enough projective modules are available to begin such constructions. Projective modules provide the lifting property that makes projective resolutions work term by term. The dual theory begins next: injective modules are introduced as the objects suited to extending maps across inclusions, completing the basic symmetry of the resolution story. # 8. Injective Modules This chapter assumes the language of rings with identity, left modules, short exact sequences, direct summands, and the projective modules introduced in the previous chapter. Injective modules are the dual objects to projective modules: projectives lift maps across surjections, while injectives extend maps across inclusions. We first isolate the extension property, then prove Baer's criterion, and finally connect injectivity with exactness of the contravariant functor $\operatorname{Hom}_R(-, I)$. The main examples come from abelian groups, where injectivity is governed by divisibility; this also provides the first supply of injective objects needed later for resolutions and derived functors. ## The Extension Problem for Modules Projective modules were introduced because a map out of a projective module can be lifted through a quotient. The dual question asks when a map into a module can be extended from a submodule to the whole module. [definition: Injective Module] Let $R$ be a ring with identity. A left $R$-module $I$ is injective if for every injective $R$-module homomorphism $\iota: A \to B$ and every $R$-module homomorphism $f: A \to I$, there exists an $R$-module homomorphism $g: B \to I$ such that $g \circ \iota = f$. [/definition] Thus an injective module has no obstruction to extending compatible maps from submodules. When $A \subset B$, the condition says that every $R$-linear map $A \to I$ extends to an $R$-linear map $B \to I$. [example: Divisibility Suggests Injectivity] Consider $\mathbb{Q}$ as a $\mathbb{Z}$-module, fix $n\ge 1$, and let $f:n\mathbb{Z}\to \mathbb{Q}$ be a group homomorphism. If $f(n)=q$, then every element of $n\mathbb{Z}$ has the form $nk$ with $k\in \mathbb{Z}$, and $\mathbb{Z}$-linearity gives \begin{align*} f(nk)=f(k\cdot n)=k f(n)=kq. \end{align*} Thus the value of $f$ is completely determined by $q$. To extend $f$ to a homomorphism $g:\mathbb{Z}\to \mathbb{Q}$, it is enough to choose $g(1)$, because then $g(k)=kg(1)$ for all $k\in\mathbb{Z}$. The condition $g|_{n\mathbb{Z}}=f$ requires \begin{align*} g(n)=f(n)=q. \end{align*} Since $g(n)=g(n\cdot 1)=n g(1)$, this becomes \begin{align*} n g(1)=q. \end{align*} In $\mathbb{Q}$ we can choose \begin{align*} g(1)=\frac{q}{n}, \end{align*} and then \begin{align*} g(nk)=nk\cdot \frac{q}{n}=kq=f(nk) \end{align*} for every $k\in\mathbb{Z}$. The divisibility of $\mathbb{Q}$ is exactly what removes the obstruction to extending maps from $n\mathbb{Z}$ to $\mathbb{Z}$. [/example] The extension property can be recognised in several equivalent ways. The point of the theorem below is to show that injectivity can be tested either by extending maps across inclusions or by splitting every overmodule, which makes it the precise dual of the standard lifting characterisations of projective modules. [quotetheorem:4553] [citeproof:4553] This result shows that injectivity is not only about extending maps; it is also about being a summand of every overmodule. The hypotheses matter: the maps are $R$-linear monomorphisms in the module category, so the quotient $M/I$ exists and splitting has its usual direct-sum meaning. The theorem does not say that every submodule of an injective module is injective, nor that a module becomes injective merely because it appears inside a large injective module. Its negative form is often more practical: if one can find a single overmodule $M$ of $I$ for which no retraction $M\to I$ exists, then $I$ is not injective. The next example gives the basic obstruction over $\mathbb{Z}$. [example: The Integers Are Not Injective] The inclusion $\mathbb{Z}\subset \mathbb{Q}$ gives the short exact sequence \begin{align*} 0 \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Q} \longrightarrow \mathbb{Q}/\mathbb{Z} \longrightarrow 0, \end{align*} where the first map is inclusion and the second sends $q$ to $q+\mathbb{Z}$. If $\mathbb{Z}$ were injective as a $\mathbb{Z}$-module, then by *Equivalent Characterisations of Injective Modules* the inclusion $\mathbb{Z}\hookrightarrow \mathbb{Q}$ would have a retraction $r:\mathbb{Q}\to\mathbb{Z}$ satisfying $r(k)=k$ for every $k\in\mathbb{Z}$. We show that no such retraction can exist. Let $r:\mathbb{Q}\to\mathbb{Z}$ be any group homomorphism. For every $n\ge 1$, \begin{align*} 1=n\cdot \frac{1}{n}, \end{align*} so $\mathbb{Z}$-linearity gives \begin{align*} r(1)=r\left(n\cdot \frac{1}{n}\right)=n\,r\left(\frac{1}{n}\right). \end{align*} Thus $r(1)$ is divisible in $\mathbb{Z}$ by every positive integer $n$. If $a=r(1)\ne 0$, take $n=|a|+1$. Since $a=n b$ for some $b\in\mathbb{Z}$, we have either $b=0$, giving $a=0$, or $|b|\ge 1$, giving \begin{align*} |a|=n|b|\ge n=|a|+1, \end{align*} which is impossible. Hence $r(1)=0$. Now for any rational number $m/n$ with $m\in\mathbb{Z}$ and $n\ge 1$, \begin{align*} 0=r(1)=r\left(n\cdot \frac{1}{n}\right)=n\,r\left(\frac{1}{n}\right), \end{align*} and since $\mathbb{Z}$ has no nonzero element killed by $n$, this gives $r(1/n)=0$. Therefore \begin{align*} r\left(\frac{m}{n}\right)=r\left(m\cdot \frac{1}{n}\right)=m\,r\left(\frac{1}{n}\right)=0. \end{align*} Every homomorphism $\mathbb{Q}\to\mathbb{Z}$ is therefore zero, while a retraction would have to satisfy $r(1)=1$. Hence $\mathbb{Z}$ is not injective as a $\mathbb{Z}$-module. [/example] The example also marks the difference between projective and injective behaviour over $\mathbb{Z}$. The module $\mathbb{Z}$ is free and hence projective, but it fails the divisibility needed for injectivity. ## Baer's Criterion The definition of injectivity asks for extensions from every submodule of every module, which is too large a class to test directly. Baer's criterion reduces the test to maps from left ideals of the ring. [quotetheorem:4203] [citeproof:4203] Baer's criterion is powerful because ideals encode all possible obstructions to extending by one new generator. The reduction is special to modules over a ring with identity: adjoining a single element $b$ to a partial domain produces the annihilator-type left ideal $J=\{r\in R: rb\in C\}$, and the element $1\in R$ records the value that the extension should assign to $b$. The criterion does not usually reduce the problem to principal ideals; that simplification occurs only for rings such as principal ideal domains. Thus over a general ring one must test all left ideals, while over $\mathbb{Z}$ the condition becomes a concrete divisibility test. [example: Checking Injectivity Over a Principal Ideal Domain] Let $D$ be an abelian group, viewed as a $\mathbb{Z}$-module. Every ideal of $\mathbb{Z}$ has the form $n\mathbb{Z}$ with $n\ge 0$, so by *Baer Criterion* it is enough to understand when maps $n\mathbb{Z}\to D$ extend to maps $\mathbb{Z}\to D$. For $n=0$, the subgroup $0\mathbb{Z}$ is the zero group, and its only homomorphism to $D$ extends to $\mathbb{Z}$ by the zero map. Now assume $n\ge 1$, and let $f:n\mathbb{Z}\to D$ be a group homomorphism. Put \begin{align*} d=f(n). \end{align*} Every element of $n\mathbb{Z}$ has the form $nk$ with $k\in\mathbb{Z}$, and $\mathbb{Z}$-linearity gives \begin{align*} f(nk)=f(k\cdot n)=k f(n)=kd. \end{align*} Thus $f$ is determined by $d$. An extension $g:\mathbb{Z}\to D$ is determined by $x=g(1)$, since for every $k\in\mathbb{Z}$, \begin{align*} g(k)=g(k\cdot 1)=k g(1)=kx. \end{align*} The condition that $g$ extend $f$ requires \begin{align*} g(n)=f(n)=d. \end{align*} But \begin{align*} g(n)=g(n\cdot 1)=n g(1)=nx, \end{align*} so the extension condition is exactly \begin{align*} nx=d. \end{align*} Conversely, if there is an element $x\in D$ with $nx=d$, define $g:\mathbb{Z}\to D$ by \begin{align*} g(k)=kx. \end{align*} Then for every $k\in\mathbb{Z}$, \begin{align*} g(nk)=nkx=k(nx)=kd=f(nk), \end{align*} so $g$ really extends $f$. Therefore Baer's criterion says that $D$ is injective exactly when every equation $nx=d$ with $n\ge 1$ and $d\in D$ has a solution in $D$, which is precisely divisibility. [/example] The previous example is usually the first complete classification of injective modules encountered in the course. To state that classification cleanly, the equation-solving property isolated by Baer's criterion needs its own name. This property will also give a concrete test for injectivity in abelian groups, replacing the abstract extension condition by the ability to divide elements by every positive integer. [definition: Divisible Abelian Group] An abelian group $D$ is divisible if for every $d \in D$ and every integer $n \ge 1$, there exists $x \in D$ such that $nx=d$. [/definition] Divisibility says that multiplication by $n$ is surjective for every $n\ge 1$. Over $\mathbb{Z}$, every ideal is generated by one integer, so Baer's criterion reduces the extension problem to solving equations of the form $nx=d$. The theorem below records the resulting classification: injective abelian groups are exactly the divisible ones. [quotetheorem:4554] [citeproof:4554] This theorem is a classification only in the category of abelian groups, equivalently $\mathbb{Z}$-modules. Its proof uses the fact that every ideal of $\mathbb{Z}$ is principal, so there is no analogous statement saying that injective modules over an arbitrary ring are just modules divisible by all elements of the ring. What it does provide is a dependable stock of injective abelian groups, and these examples become the targets used to build concrete injective resolutions later. [example: Rational and Prüfer-Type Examples] Fix $n\ge 1$. For any $q\in\mathbb{Q}$, choose $x=q/n\in\mathbb{Q}$. Then \begin{align*} nx=n\cdot \frac{q}{n}=q, \end{align*} so multiplication by $n$ is surjective on $\mathbb{Q}$. The same calculation works in the quotient $\mathbb{Q}/\mathbb{Z}$. Given a class $q+\mathbb{Z}$, choose \begin{align*} x=\frac{q}{n}+\mathbb{Z}. \end{align*} Using the quotient group operation, multiplication by $n$ gives \begin{align*} nx &=n\left(\frac{q}{n}+\mathbb{Z}\right)\\ &=n\cdot \frac{q}{n}+\mathbb{Z}\\ &=q+\mathbb{Z}. \end{align*} Thus multiplication by $n$ is surjective on both $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$ for every $n\ge 1$, so both groups are divisible. By *Divisible Abelian Groups Are Injective*, both $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$ are injective $\mathbb{Z}$-modules. [/example] The quotient $\mathbb{Q}/\mathbb{Z}$ is especially important because it is torsion but still injective. It later becomes a standard target for duality constructions on abelian groups. ## Exactness of $\operatorname{Hom}_R(-, I)$ The functor $\operatorname{Hom}_R(-, I)$ is contravariant: a map $A\to B$ induces a map $\operatorname{Hom}_R(B,I)\to \operatorname{Hom}_R(A,I)$ by precomposition. Injectivity is precisely the condition that this left exact functor becomes exact. [quotetheorem:4555] [citeproof:4555] This theorem is the formal bridge from injective modules to derived functors. The first two nonzero terms of the Hom sequence are exact for every target $I$; injectivity is needed only at the final surjectivity statement, where maps $A\to I$ must extend across $A\hookrightarrow B$. Without injectivity this last map can fail to be onto, so the theorem is not a formal property of contravariant Hom alone. Chapter 10 will use this exactness property to build injective resolutions, and the following example isolates the point where non-injectivity breaks exactness. [example: Failure of Exactness for $I=\mathbb{Z}$] Apply $\operatorname{Hom}_{\mathbb{Z}}(-,\mathbb{Z})$ to the short exact sequence \begin{align*} 0 \longrightarrow \mathbb{Z} \xrightarrow{\times 2} \mathbb{Z} \longrightarrow \mathbb{Z}/2\mathbb{Z} \longrightarrow 0. \end{align*} Contravariance reverses the arrows, so the relevant map \begin{align*} \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z}) \longrightarrow \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z}) \end{align*} sends a homomorphism $\varphi:\mathbb{Z}\to\mathbb{Z}$ to the composite \begin{align*} \mathbb{Z}\xrightarrow{\times 2}\mathbb{Z}\xrightarrow{\varphi}\mathbb{Z}. \end{align*} Identify $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z})$ with $\mathbb{Z}$ by sending $\varphi$ to $\varphi(1)$. If $\varphi(1)=a$, then for every $k\in\mathbb{Z}$, \begin{align*} (\varphi\circ(\times 2))(k) &=\varphi(2k)\\ &=\varphi((2k)\cdot 1)\\ &=(2k)\varphi(1)\\ &=2ka. \end{align*} In particular, \begin{align*} (\varphi\circ(\times 2))(1)=2a. \end{align*} Thus, under the identification $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z})\cong\mathbb{Z}$, the induced map is \begin{align*} \mathbb{Z}\longrightarrow\mathbb{Z}, \qquad a\longmapsto 2a. \end{align*} This map is not surjective: the integer $1$ is not equal to $2a$ for any $a\in\mathbb{Z}$, since $2a$ is always even. Equivalently, the identity map $\operatorname{id}_{\mathbb{Z}}$ has value $\operatorname{id}_{\mathbb{Z}}(1)=1$, so it cannot be the composite $\varphi\circ(\times 2)$ for any endomorphism $\varphi$ of $\mathbb{Z}$. Therefore the final map in the Hom sequence fails to be onto, so $\operatorname{Hom}_{\mathbb{Z}}(-,\mathbb{Z})$ is not exact; this is the exactness failure corresponding to the non-injectivity of $\mathbb{Z}$. [/example] Exactness also explains the duality with projective modules. A module $P$ is projective precisely when $\operatorname{Hom}_R(P,-)$ is exact, while a module $I$ is injective precisely when $\operatorname{Hom}_R(-,I)$ is exact. ## Embedding Modules into Injectives Projective resolutions exist because free modules are plentiful: every module is a quotient of a free module by sending basis elements to chosen generators. The injective side is less visible, since there is no comparably elementary construction that embeds an arbitrary module into a visibly injective free-like object. For example, the natural embedding $\mathbb{Z}\hookrightarrow\mathbb{Q}$ works only because divisibility has already identified $\mathbb{Q}$ as injective. The general theory needs the dual supply statement, namely that every module embeds into some injective module. [quotetheorem:4556] [citeproof:4556] For this course, the essential consequence is the existence of injective resolutions. Starting from $M\hookrightarrow I^0$, the cokernel embeds into another injective $I^1$, and iteration gives a cochain complex beginning at $M$. The statement is the dual supply theorem to the existence of projective resolutions: it guarantees that the injective side has enough objects to support derived functor constructions. [definition: Injective Resolution] An injective resolution of a left $R$-module $M$ is an exact sequence \begin{align*} 0 \longrightarrow M \longrightarrow I^0 \longrightarrow I^1 \longrightarrow I^2 \longrightarrow \cdots \end{align*} where each $I^n$ is an injective left $R$-module. [/definition] The indexing is cohomological because injective resolutions are usually used to derive left exact covariant functors or contravariant functors in the appropriate variable. The initial copy of $M$ is not itself required to be injective. [example: Beginning an Injective Resolution of $\mathbb{Z}$] The map $\mathbb{Z}\hookrightarrow \mathbb{Q}$ is the usual inclusion, so it is injective because an integer that maps to $0\in\mathbb{Q}$ is already $0\in\mathbb{Z}$. The group $\mathbb{Q}$ is divisible: for $q\in\mathbb{Q}$ and $n\ge 1$, the element $q/n\in\mathbb{Q}$ satisfies \begin{align*} n\left(\frac{q}{n}\right)=q. \end{align*} Hence $\mathbb{Q}$ is injective as a $\mathbb{Z}$-module by *Divisible Abelian Groups Are Injective*. Let $\pi:\mathbb{Q}\to \mathbb{Q}/\mathbb{Z}$ be the quotient map, $\pi(q)=q+\mathbb{Z}$. This map is surjective because every element of $\mathbb{Q}/\mathbb{Z}$ has the form $q+\mathbb{Z}=\pi(q)$. Its kernel is \begin{align*} \ker(\pi) &=\{q\in\mathbb{Q}:q+\mathbb{Z}=\mathbb{Z}\}\\ &=\{q\in\mathbb{Q}:q\in\mathbb{Z}\}\\ &=\mathbb{Z}. \end{align*} Therefore \begin{align*} 0 \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Q} \xrightarrow{\pi} \mathbb{Q}/\mathbb{Z} \longrightarrow 0 \end{align*} is exact. The quotient $\mathbb{Q}/\mathbb{Z}$ is also divisible. Given $q+\mathbb{Z}\in \mathbb{Q}/\mathbb{Z}$ and $n\ge 1$, choose $x=\frac{q}{n}+\mathbb{Z}$. Then \begin{align*} nx &=n\left(\frac{q}{n}+\mathbb{Z}\right)\\ &=n\cdot \frac{q}{n}+\mathbb{Z}\\ &=q+\mathbb{Z}. \end{align*} So $\mathbb{Q}/\mathbb{Z}$ is injective by *Divisible Abelian Groups Are Injective*. Thus the displayed exact sequence is an injective resolution of $\mathbb{Z}$ of length one: after the initial module $\mathbb{Z}$, both terms are injective. [/example] Injective modules complete the basic symmetry begun with projective modules. Projectives are adapted to quotients, injectives are adapted to submodules, and both provide the resolutions that will be used to define and compute derived functors. Projective and injective modules now give the two complementary ways to resolve an arbitrary module. The next chapter focuses on projective resolutions, using them to replace a module by an exact complex of projectives in a form suited to derived constructions. # 9. Projective Resolutions Projective resolutions are the mechanism that replaces an arbitrary module by a chain complex built from projective modules. The replacement is exact except at the target module, so it records the module while making lifting arguments possible term by term. In the previous chapters, homotopy explained when two maps of complexes should be regarded as the same; here that language becomes essential because resolutions are unique only up to chain homotopy. ## Why Resolutions Are Needed Many functors in algebra preserve some exact sequences but not all of them. The role of a projective resolution is to present a module by projective objects in every degree, so that maps out of the resolution can be lifted through surjections and then compared by homotopy. Before defining the resolution itself, we isolate the ambient object: an augmented complex ending in the module being resolved. [definition: Augmented Chain Complex] Let $R$ be a ring and let $M$ be a left $R$-module. An augmented chain complex over $M$ is a chain complex of left $R$-modules \begin{align*} \cdots \xrightarrow{d_3} P_2 \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \end{align*} together with an $R$-module homomorphism $\varepsilon: P_0 \to M$ such that $\varepsilon d_1 = 0$. [/definition] The augmentation lets us append $M$ at degree $-1$ and read the condition $\varepsilon d_1=0$ as the statement that the displayed sequence continues as a complex. To turn such an augmented complex into a useful replacement for $M$, two more requirements are needed: the sequence should be exact, and the terms before $M$ should have the lifting property supplied by projective modules. We need this strengthened object because later comparisons and derived constructions require a replacement of $M$ by modules that lift maps, not merely a complex ending at $M$. [definition: Projective Resolution] Let $R$ be a ring and let $M$ be a left $R$-module. A projective resolution of $M$ is an augmented chain complex \begin{align*} \cdots \xrightarrow{d_3} P_2 \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \xrightarrow{\varepsilon} M \to 0 \end{align*} such that each $P_n$ is a projective left $R$-module and the displayed sequence is exact. [/definition] Thus a projective resolution is acyclic after the target module is included as the augmented term. The adjective projective is not cosmetic: it is the lifting property of the $P_n$ that makes comparison maps and homotopies exist. [example: Length-One Resolution Of A Cyclic Quotient] Let $R=\mathbb Z$ and let $M=\mathbb Z/n\mathbb Z$ with $n\ge 1$. Define $d_1:\mathbb Z\to\mathbb Z$ by $d_1(a)=na$, and let $\varepsilon:\mathbb Z\to \mathbb Z/n\mathbb Z$ be the quotient map $\varepsilon(b)=b+n\mathbb Z$. The composite is zero, since \begin{align*} (\varepsilon d_1)(a) &=\varepsilon(na) \\ &=na+n\mathbb Z \\ &=0+n\mathbb Z. \end{align*} Also $d_1$ is injective: if $d_1(a)=0$, then $na=0$ in $\mathbb Z$, and because $n\ge 1$ this forces $a=0$. The image of $d_1$ is exactly $n\mathbb Z$, since \begin{align*} \operatorname{im} d_1 &=\{d_1(a):a\in\mathbb Z\} \\ &=\{na:a\in\mathbb Z\} \\ &=n\mathbb Z. \end{align*} The kernel of $\varepsilon$ is also $n\mathbb Z$, because \begin{align*} b\in\ker\varepsilon &\Longleftrightarrow b+n\mathbb Z=0+n\mathbb Z \\ &\Longleftrightarrow b\in n\mathbb Z. \end{align*} Thus the sequence \begin{align*} 0 \to \mathbb Z \xrightarrow{\cdot n} \mathbb Z \xrightarrow{\varepsilon} \mathbb Z/n\mathbb Z \to 0 \end{align*} is exact. Both copies of $\mathbb Z$ are free, hence projective, as $\mathbb Z$-modules, so this is a projective resolution of $\mathbb Z/n\mathbb Z$ concentrated in degrees $1$ and $0$. This example shows that a familiar presentation by one generator and one relation is already a resolution when the relation module is free. [/example] A resolution may be finite, as in the cyclic quotient example, or infinite. Once resolutions are allowed to continue through many degrees, we need a way to record where the nonzero projective terms stop. The length of a projective resolution is the first numerical measure of this size, and it becomes a homological measure of how far the module is from being projective. [definition: Length Of A Projective Resolution] A projective resolution $P_\bullet \to M$ has length at most $m$ if $P_n=0$ for all $n>m$. [/definition] The length convention makes projective modules the modules with resolutions of length $0$: the augmentation $P_0\to M$ can be chosen to be an isomorphism. Later courses package the minimum possible length as projective dimension. ## Constructing Resolutions From Successive Kernels The natural construction problem is the following: given a module $M$, how do we build a resolution without already knowing its relations in all degrees? The answer is iterative. Choose a projective module mapping onto $M$, take the kernel, choose a projective module mapping onto that kernel, and continue. [quotetheorem:4557] [citeproof:4557] This construction is useful because it reduces the global problem of exactness to a repeated presentation problem. The hypothesis that we are working in the full category of modules is doing real work: every module is a quotient of a free module, while smaller module settings may fail to contain enough projective objects for such a construction. The theorem also does not say that the resolution is finite, canonical, or minimal; it only guarantees some free resolution, possibly with very large modules in every degree. Its importance is that the successive kernels are the syzygies of $M$, and applying functors to these free resolutions is the starting point for invariants such as $\operatorname{Tor}$, $\operatorname{Ext}$, and projective dimension. [example: Standard Presentation Resolution] Suppose $R$ is a ring and a left $R$-module $M$ has a finite presentation \begin{align*} R^m \xrightarrow{A} R^n \xrightarrow{\pi} M \to 0, \end{align*} where $\pi$ is surjective and the sequence is exact at $R^n$. Exactness at $R^n$ means \begin{align*} \operatorname{im} A=\ker \pi. \end{align*} Since $R^m$ and $R^n$ are free left $R$-modules, the displayed sequence gives the first two free terms of a resolution: \begin{align*} R^m \xrightarrow{A} R^n \xrightarrow{\pi} M \to 0. \end{align*} To extend it one step, assume $\ker A$ is finitely generated. Choose generators $k_1,\dots,k_\ell$ for $\ker A$ and define an $R$-linear map \begin{align*} B:R^\ell\to R^m \end{align*} by \begin{align*} B(e_i)=k_i \end{align*} on the standard basis vectors $e_i$ of $R^\ell$. For an element $\sum_{i=1}^{\ell} r_i e_i\in R^\ell$, we have \begin{align*} (AB)\left(\sum_{i=1}^{\ell} r_i e_i\right) &=A\left(B\left(\sum_{i=1}^{\ell} r_i e_i\right)\right) \\ &=A\left(\sum_{i=1}^{\ell} r_i B(e_i)\right) \\ &=A\left(\sum_{i=1}^{\ell} r_i k_i\right) \\ &=\sum_{i=1}^{\ell} r_i A(k_i) \\ &=\sum_{i=1}^{\ell} r_i\cdot 0 \\ &=0, \end{align*} because each $k_i$ lies in $\ker A$. Thus $\operatorname{im}B\subseteq\ker A$. Conversely, if $u\in\ker A$, then $u$ is an $R$-linear combination of the chosen generators, so \begin{align*} u=\sum_{i=1}^{\ell} r_i k_i =B\left(\sum_{i=1}^{\ell} r_i e_i\right), \end{align*} and hence $u\in\operatorname{im}B$. Therefore \begin{align*} \operatorname{im}B=\ker A. \end{align*} So the extended sequence \begin{align*} R^\ell \xrightarrow{B} R^m \xrightarrow{A} R^n \xrightarrow{\pi} M \to 0 \end{align*} is exact at $R^m$, exact at $R^n$, and exact at $M$ because $\pi$ is surjective. Repeating the same construction with $\ker B$, then with the next kernel, produces a free resolution; the successive matrices encode generators, relations, relations among relations, and higher syzygies. [/example] The previous example is the computational face of the successive-kernel method. In commutative algebra, these matrices become the input for syzygies and homological invariants. [example: Periodic Resolution Over Dual Numbers] Let $k$ be a field and let $R=k[x]/(x^2)$. Write the class of $x$ in $R$ again as $x$, so $x^2=0$ in $R$. Every element of $R$ has a unique form $a+bx$ with $a,b\in k$. Regard $k$ as an $R$-module through the quotient map \begin{align*} \varepsilon:R\to R/(x)\cong k, \qquad \varepsilon(a+bx)=a. \end{align*} The differential in every positive degree is multiplication by $x$. For $a+bx\in R$, \begin{align*} x(a+bx) &=ax+bx^2 \\ &=ax+b\cdot 0 \\ &=ax. \end{align*} Thus \begin{align*} \operatorname{im}(\cdot x) &=\{x(a+bx):a,b\in k\} \\ &=\{ax:a\in k\} \\ &=(x). \end{align*} The kernel is the same ideal, because \begin{align*} a+bx\in\ker(\cdot x) &\Longleftrightarrow x(a+bx)=0 \\ &\Longleftrightarrow ax=0 \\ &\Longleftrightarrow a=0 \\ &\Longleftrightarrow a+bx=bx\in (x). \end{align*} Also \begin{align*} \varepsilon(x(a+bx)) &=\varepsilon(ax) \\ &=0, \end{align*} so $\operatorname{im}(\cdot x)\subseteq\ker\varepsilon$. Conversely, \begin{align*} a+bx\in\ker\varepsilon &\Longleftrightarrow \varepsilon(a+bx)=0 \\ &\Longleftrightarrow a=0 \\ &\Longleftrightarrow a+bx=bx\in(x). \end{align*} Therefore $\ker\varepsilon=(x)=\operatorname{im}(\cdot x)$, and at every earlier copy of $R$ we have $\ker(\cdot x)=(x)=\operatorname{im}(\cdot x)$. Hence \begin{align*} \cdots \xrightarrow{\cdot x} R \xrightarrow{\cdot x} R \xrightarrow{\cdot x} R \xrightarrow{\varepsilon} k \to 0 \end{align*} is exact. Each copy of $R$ is free of rank one as an $R$-module, so this is a free resolution of $k$ which repeats with period one. The example shows that even the two-dimensional $k$-algebra $k[x]/(x^2)$ can force an infinite resolution. [/example] The periodic resolution is also a warning against expecting resolutions to terminate. Infinite resolutions are not an artefact of poor choices; over many rings they are unavoidable. ## Lifting Maps Between Modules Once modules have been replaced by projective resolutions, a map between the original modules should lift to a chain map between the chosen resolutions. The point is that the lift is constructed degree by degree, using projectivity against the surjections onto kernels. [quotetheorem:4558] [citeproof:4558] The theorem says that a module map can be studied after choosing projective replacements. Both projectivity and exactness are essential: if $P_0$ were not projective, a map $P_0\to N$ need not lift through a surjection $Q_0\to N$; for instance, the identity map $\mathbb Z/2\mathbb Z\to\mathbb Z/2\mathbb Z$ does not lift through $\mathbb Z/4\mathbb Z\to\mathbb Z/2\mathbb Z$ as a homomorphism of abelian groups. Exactness is equally necessary because the induction lifts into kernels using the equality $\ker d_n^Q=\operatorname{im}d_{n+1}^Q$. The theorem does not make the lift unique or functorial on the nose; it only guarantees existence for the chosen resolutions. Since the construction involves choices at every degree, the next question is whether different choices produce genuinely different maps of complexes. [definition: Chain Homotopy Between Lifted Maps] Let $g,h:P_\bullet\to Q_\bullet$ be chain maps between chain complexes. A chain homotopy from $g$ to $h$ is a family of $R$-module homomorphisms $s_n:P_n\to Q_{n+1}$ such that \begin{align*} g_n-h_n=d_{n+1}^Q s_n+s_{n-1}d_n^P \end{align*} for all $n\ge 0$, with the convention $s_{-1}=0$ at the augmented end. [/definition] Chain homotopy is the right equality relation for maps between resolutions. It preserves the induced maps on homology, and resolutions have no positive-degree homology, so the obstruction to uniqueness disappears through another induction. [quotetheorem:4559] [citeproof:4559] This uniqueness theorem is why constructions made using projective resolutions do not depend on the chosen resolution, at least after passing to homotopy-invariant data. The assumptions cannot simply be dropped: projectivity is used to solve each lifting problem for the homotopy maps $s_n$, and exactness of $Q_\bullet$ is used to convert the defect at degree $n$ into something hit by $d_{n+1}^Q$. Without those hypotheses, two chain maps can induce the same map on the augmented module but fail to be connected by this recursive homotopy construction. The theorem also does not say that the lifted maps are equal, nor that there is a preferred homotopy; it says only that their difference vanishes in the homotopy category. In later homological algebra, this is exactly what makes derived functors well-defined, because $\operatorname{Hom}_R(P_\bullet,-)$ and tensor constructions depend only on the homotopy class of the chosen comparison map. [remark: Resolutions As Replacements] A projective resolution should be read as a replacement $P_\bullet\to M$ rather than as extra structure on $M$. The comparison theorem says that maps of modules lift to maps of replacements, and homotopy uniqueness says that the lift is unique in the homotopy category. This is the first appearance of the derived viewpoint: replace objects by better objects, then compute using maps between the replacements. [/remark] ## What This Chapter Establishes Projective resolutions exist because free modules are plentiful, and they can be built by repeatedly resolving the current kernel. The comparison theorem shows that resolutions are functorial up to a controlled choice, while homotopy uniqueness shows that this choice does not affect homotopy-invariant constructions. These facts are the projective half of the resolution technology; the next parallel development replaces maps out of projectives by maps into injectives. Projective resolutions have supplied the main computational framework of the course, and the dual injective theory completes the picture. The final chapter develops injective resolutions and then ties the whole subject together, showing how both kinds of resolutions organize the homological algebra built so far. # 10. Injective Resolutions and Course Synthesis ## Building Injective Resolutions from Cokernels The projective construction starts with a surjection onto the module being studied. The dual problem is to start with a monomorphism out of the module, so the first question is: when can every module be embedded into a module with enough extension properties? As in Chapter 8, an injective left $R$-module is one for which maps into it extend across monomorphisms. This is the exact dual of projectivity from Chapter 7, where maps out of a projective module lift across epimorphisms. The construction of injective resolutions depends on the ambient category having enough injective objects. [definition: Enough Injectives for Modules] The category of left $R$-modules has enough injectives if every left $R$-module $M$ admits an injective embedding \begin{align*} M\hookrightarrow I \end{align*} into an injective left $R$-module $I$. [/definition] In the category of modules over a ring, enough injectives is a theorem rather than an extra assumption. Chapter 8 introduced injective modules through extension properties and concrete abelian-group examples; here we use the existence theorem to construct resolutions. [quotetheorem:4560] [citeproof:4560] The hypothesis that we are working in the whole module category is essential. A smaller module setting can fail to contain enough injective objects: among finitely generated abelian groups, the only divisible finitely generated group is $0$, so nonzero objects such as $\mathbb Z$ cannot embed into an injective object that remains finitely generated. Thus the theorem uses the availability of large modules. Once embeddings into injectives are available, the next task is to iterate them without losing exactness. The cokernel replaces the kernel from the projective construction. [definition: Injective Resolution] Let $R$ be a ring and let $M$ be a left $R$-module. An injective resolution of $M$ is an exact cochain complex \begin{align*} 0 \longrightarrow M \xrightarrow{\varepsilon} I^0 \xrightarrow{d^0} I^1 \xrightarrow{d^1} I^2 \xrightarrow{d^2} \cdots \end{align*} in which every $I^q$ is an injective left $R$-module. [/definition] Exactness at $I^0$ says that $\ker d^0 = \operatorname{im}\varepsilon$, and exactness at later terms says that $\ker d^q = \operatorname{im} d^{q-1}$. Thus the module $M$ appears as the degree $0$ cohomology of the augmented complex, while all higher cohomology vanishes. [explanation: Existence Of Injective Resolutions For Modules] Every left $R$-module $M$ admits an injective resolution \begin{align*} 0 \longrightarrow M \longrightarrow I^0 \longrightarrow I^1 \longrightarrow I^2 \longrightarrow \cdots \end{align*} with each $I^q$ an injective left $R$-module. [/explanation] The construction is canonical only up to many choices, and the statement says nothing about finiteness, uniqueness, or minimality of the injective terms. Its hypothesis is the module-category supply of enough injective modules: the first arrow $M \to I^0$ must exist before the induction can even begin. The category of finitely generated abelian groups gives a concrete warning about changing the ambient category, since $\mathbb Z$ is finitely generated but cannot embed into a nonzero injective object inside that smaller category. The point of the comparison theorem later in the chapter is that the choices made in the full module category do not matter after passing to homotopy. [example: The Basic Injective Resolution of Integers] For every integer $n\ge 1$ and every $q\in \mathbb Q$, the element $q/n\in \mathbb Q$ satisfies $n(q/n)=q$, so $\mathbb Q$ is divisible. The quotient $\mathbb Q/\mathbb Z$ is also divisible: if $x=q+\mathbb Z$ and $n\ge 1$, then \begin{align*} n\left(\frac{q}{n}+\mathbb Z\right) = q+\mathbb Z = x. \end{align*} By *Divisible Abelian Groups Are Injective*, both $\mathbb Q$ and $\mathbb Q/\mathbb Z$ are injective abelian groups. Let $\iota:\mathbb Z\to\mathbb Q$ be the usual inclusion and let $\pi:\mathbb Q\to\mathbb Q/\mathbb Z$ be the quotient map. Then $\iota$ is injective, $\pi$ is surjective, and \begin{align*} \ker \pi &= \{q\in \mathbb Q : q+\mathbb Z=\mathbb Z\} \\ &= \{q\in \mathbb Q : q\in \mathbb Z\} \\ &= \mathbb Z \\ &= \operatorname{im}\iota . \end{align*} Thus \begin{align*} 0 \longrightarrow \mathbb Z \xrightarrow{\iota} \mathbb Q \xrightarrow{\pi} \mathbb Q/\mathbb Z \longrightarrow 0 \end{align*} is exact, with injective terms $\mathbb Q$ and $\mathbb Q/\mathbb Z$. Hence it is an injective resolution of $\mathbb Z$ in the category of abelian groups. The example shows that injective resolutions need not resemble free resolutions: the terms are large divisible groups, and their extension property is what matters. [/example] The same cokernel method applies to finite cyclic groups, although the injective modules that appear are usually larger than the original module. [example: First Steps for a Cyclic Group] Let $n\ge 2$, and define $\phi:\mathbb Z/n\mathbb Z\to \mathbb Q/\mathbb Z$ by \begin{align*} \phi(\bar{k})=\frac{k}{n}+\mathbb Z. \end{align*} If $\bar{k}=\bar{\ell}$ in $\mathbb Z/n\mathbb Z$, then $k-\ell=an$ for some $a\in\mathbb Z$, so \begin{align*} \frac{k}{n}-\frac{\ell}{n} = \frac{k-\ell}{n} = a \in \mathbb Z. \end{align*} Thus $k/n+\mathbb Z=\ell/n+\mathbb Z$, so $\phi$ is well-defined. It is a homomorphism because \begin{align*} \phi(\bar{k}+\bar{\ell}) &= \phi(\overline{k+\ell}) \\ &= \frac{k+\ell}{n}+\mathbb Z \\ &= \left(\frac{k}{n}+\mathbb Z\right)+\left(\frac{\ell}{n}+\mathbb Z\right) \\ &= \phi(\bar{k})+\phi(\bar{\ell}). \end{align*} Its kernel is \begin{align*} \ker \phi &= \left\{\bar{k}\in \mathbb Z/n\mathbb Z : \frac{k}{n}+\mathbb Z=\mathbb Z\right\} \\ &= \left\{\bar{k}\in \mathbb Z/n\mathbb Z : \frac{k}{n}\in \mathbb Z\right\} \\ &= \{\bar{k}\in \mathbb Z/n\mathbb Z : n\mid k\} \\ &= \{\bar{0}\}, \end{align*} so $\phi$ is injective. Its image is the cyclic subgroup generated by $1/n+\mathbb Z$, since \begin{align*} \operatorname{im}\phi &= \left\{\frac{k}{n}+\mathbb Z : k\in\mathbb Z\right\} \\ &= \left\langle \frac{1}{n}+\mathbb Z\right\rangle . \end{align*} Let \begin{align*} C= (\mathbb Q/\mathbb Z)\big/\left\langle \frac{1}{n}+\mathbb Z\right\rangle . \end{align*} The quotient map $\pi:\mathbb Q/\mathbb Z\to C$ has kernel $\operatorname{im}\phi$, so \begin{align*} 0\longrightarrow \mathbb Z/n\mathbb Z \xrightarrow{\phi} \mathbb Q/\mathbb Z \xrightarrow{\pi} C \longrightarrow 0 \end{align*} is exact. The group $C$ is divisible: if $m\ge 1$ and $y=(q+\mathbb Z)+\operatorname{im}\phi\in C$, then \begin{align*} m\left(\left(\frac{q}{m}+\mathbb Z\right)+\operatorname{im}\phi\right) = (q+\mathbb Z)+\operatorname{im}\phi = y. \end{align*} It is also torsion: if $q=a/b$ with $a\in\mathbb Z$ and $b\ge 1$, then \begin{align*} b\big((q+\mathbb Z)+\operatorname{im}\phi\big) &= (a+\mathbb Z)+\operatorname{im}\phi \\ &= \mathbb Z+\operatorname{im}\phi \\ &= 0. \end{align*} By *Divisible Abelian Groups Are Injective*, both $\mathbb Q/\mathbb Z$ and $C$ are injective abelian groups. Thus one may take $I^0=\mathbb Q/\mathbb Z$ and $I^1=C$, giving the injective resolution \begin{align*} 0 \longrightarrow \mathbb Z/n\mathbb Z \xrightarrow{\phi} \mathbb Q/\mathbb Z \xrightarrow{\pi} C \longrightarrow 0. \end{align*} This is the cokernel construction in its first nontrivial form: embed the module into an injective group, identify the cokernel explicitly, and continue only when the cokernel has not already landed in an injective group. [/example] ## Comparing Injective Resolutions A resolution is useful only if maps of modules can be lifted to maps of resolutions. For injective resolutions the lifting proceeds to the right, using injectivity to extend maps across the monomorphisms that arise from exactness. [definition: Morphism of Augmented Injective Resolutions] Let $f: M \to N$ be a homomorphism of left $R$-modules. Let $I^\bullet$ be an injective resolution of $M$ and let $J^\bullet$ be an injective resolution of $N$. A morphism of augmented injective resolutions over $f$ is a cochain map $\alpha^\bullet: I^\bullet \to J^\bullet$ together with $f$ in degree $-1$ such that the square \begin{align*} \begin{array}{ccc} M & \xrightarrow{\varepsilon_I} & I^0 \\ f\downarrow & & \downarrow\alpha^0 \\ N & \xrightarrow{\varepsilon_J} & J^0 \end{array} \end{align*} commutes. [/definition] The degree $0$ square is the starting constraint. After that, the obstruction to defining $\alpha^{q+1}$ lies in making a partially defined map extend from a submodule of $I^{q+1}$ into $J^{q+1}$, and injectivity provides precisely that extension. [quotetheorem:4561] [citeproof:4561] The theorem says that the resolution process is functorial up to choices, but both injectivity and exactness are doing real work. Without injectivity of the target terms, the extension step can fail: in abelian groups, the homomorphism $2\mathbb Z \to \mathbb Z$ sending $2$ to $1$ does not extend across the inclusion $2\mathbb Z \hookrightarrow \mathbb Z$, since an extension $h: \mathbb Z \to \mathbb Z$ would require $2h(1)=1$. Without exactness of the source resolution, the composite into $J^{q+1}$ need not vanish on the required kernel, so it may not factor through the image where the extension argument starts. The choices are controlled by homotopy, which is the correct equivalence relation for complexes from the earlier chapters. [definition: Cochain Homotopy] Let $A^\bullet$ and $B^\bullet$ be cochain complexes of left $R$-modules, and let $\alpha^\bullet,\beta^\bullet: A^\bullet \to B^\bullet$ be cochain maps. A cochain homotopy from $\alpha^\bullet$ to $\beta^\bullet$ is a family of homomorphisms $s^q: A^q \to B^{q-1}$ such that \begin{align*} \alpha^q - \beta^q = d_B^{q-1}s^q + s^{q+1}d_A^q \end{align*} for every $q$. [/definition] Homotopic cochain maps induce the same map on cohomology, so homotopy uniqueness is strong enough for all later uses of resolutions. The remaining question is whether the arbitrary choices in an injective comparison map are unique up to this relation. That is the needed stability statement: different embeddings into injective modules should give equivalent information after passing to homotopy classes. [quotetheorem:4562] [citeproof:4562] As a consequence, a module homomorphism determines a well-defined homotopy class of maps between any chosen injective resolutions. The statement is limited to maps over the same module homomorphism and to genuine injective resolutions; for arbitrary complexes, two cochain maps with the same effect at an augmentation need not be homotopic. Exactness is used to turn vanishing on kernels into factorisation through images, while injectivity is used to extend the partially defined homotopy maps. The same failure seen in the inclusion $2\mathbb Z\hookrightarrow\mathbb Z$ shows why a non-injective target term cannot support the argument in general. This is the injective counterpart of the projective comparison theorem, and it is the main reason resolutions can be used without remembering the arbitrary choices made during construction. [remark: Direction of the Argument] Projective comparison lifts maps through epimorphisms, so it uses kernels and proceeds through surjective steps. Injective comparison extends maps across monomorphisms, so it uses cokernels and proceeds through inclusion steps. The proofs are formally dual, but writing down the first two degrees is often the best way to avoid reversing a differential in the wrong direction. [/remark] ## What Resolutions Contribute to Later Derived Constructions The course stops before developing the full theory of derived functors, but the reason for building resolutions is already visible. A functor that is not exact can be applied to a resolution, and the resulting homology or cohomology measures the failure of exactness. [explanation: Why Resolutions Replace Modules] A module $M$ by itself may be too small to expose how a functor behaves on exact sequences. A projective resolution replaces $M$ by a complex made of modules that lift maps well, while an injective resolution replaces $M$ by a complex made of modules that extend maps well. The replacement is exact away from $M$, so it retains the original module while giving a longer object on which functors can be tested degree by degree. The comparison theorems are what make this replacement meaningful. If different resolutions produced unrelated complexes after applying a functor, then computations would depend on arbitrary choices. Homotopy uniqueness ensures that the induced constructions are independent of the selected resolution up to the equivalence relation already known to preserve homology. [/explanation] This viewpoint explains the parallel roles of projective and injective resolutions. Projective resolutions are suited to covariant right exact functors and contravariant left exact functors, while injective resolutions are suited to covariant left exact functors. The precise derived functor definitions belong to the next course, but the input data are exactly the resolutions and homotopy comparison results established here. [quotetheorem:4563] [citeproof:4563] This principle is the technical bridge from the present course to the next stage of homological algebra. It says that computations may begin with any convenient injective resolution, just as projective computations may begin with any convenient projective resolution. The hypotheses cannot be weakened to arbitrary complexes replacing $M$: a non-exact complex can have extra cohomology, and a non-injective complex may not admit comparison maps in either direction. For example, replacing $\mathbb Z$ by the cochain complex concentrated in degree $0$ at $\mathbb Z$ is not an injective resolution in abelian groups, and there is no reason for it to be homotopy equivalent to the genuine resolution $\mathbb Q\to \mathbb Q/\mathbb Z$. The theorem is therefore independence among valid injective resolutions, not independence from the choice of any complex that happens to contain the module. [example: Acyclicity After Forgetting the Augmentation] Let $\iota:\mathbb Z\to\mathbb Q$ be the inclusion $\iota(m)=m$, and let $\pi:\mathbb Q\to\mathbb Q/\mathbb Z$ be the quotient map $\pi(q)=q+\mathbb Z$. For the non-augmented cochain complex \begin{align*} \mathbb Q \xrightarrow{\pi} \mathbb Q/\mathbb Z, \end{align*} with $\mathbb Q$ placed in degree $0$, the degree $0$ cohomology is \begin{align*} H^0 &= \ker(\pi:\mathbb Q\to\mathbb Q/\mathbb Z) \\ &= \{q\in\mathbb Q : q+\mathbb Z=\mathbb Z\} \\ &= \{q\in\mathbb Q : q\in\mathbb Z\} \\ &= \mathbb Z. \end{align*} There is no incoming differential in this non-augmented complex, so no quotient by boundaries appears in degree $0$. After including the augmentation, the sequence is \begin{align*} 0 \longrightarrow \mathbb Z \xrightarrow{\iota} \mathbb Q \xrightarrow{\pi} \mathbb Q/\mathbb Z \longrightarrow 0. \end{align*} Exactness at $\mathbb Z$ holds because \begin{align*} \ker \iota &= \{m\in\mathbb Z : \iota(m)=0\} \\ &= \{m\in\mathbb Z : m=0\} \\ &= 0. \end{align*} Exactness at $\mathbb Q$ holds because \begin{align*} \operatorname{im}\iota &= \{\iota(m):m\in\mathbb Z\} \\ &= \{m:m\in\mathbb Z\} \\ &= \mathbb Z \\ &= \ker\pi. \end{align*} Exactness at $\mathbb Q/\mathbb Z$ holds because every element has the form $q+\mathbb Z=\pi(q)$, so \begin{align*} \operatorname{im}\pi=\mathbb Q/\mathbb Z. \end{align*} Thus forgetting the augmentation turns the stored copy of $\mathbb Z$ into degree $0$ cohomology, while the augmented exact sequence records that $\mathbb Q\to\mathbb Q/\mathbb Z$ is a replacement for $\mathbb Z$ rather than a complex meant to have zero cohomology at $\mathbb Q$. [/example] ## Synthesis of the Course The main thread of the course has been the replacement of individual module calculations by structured calculations with complexes. The final synthesis is to identify which ideas survive every change of resolution and which are artifacts of a chosen model. [explanation: The Core Pattern] Chain complexes organize algebraic data by placing maps in sequence and asking whether consecutive maps compose to zero. Homology measures the difference between cycles and boundaries, so exact complexes are precisely those with no internal homological obstruction. Short exact sequences of complexes produce long exact sequences in homology, and diagram lemmas such as the snake lemma provide the local mechanism behind these connecting maps. Homotopy then weakens equality of maps to a relation that preserves homology. Mapping cones, contractible complexes, and homotopy equivalences show that complexes should often be compared by homotopy type rather than termwise isomorphism. Projective and injective resolutions are the two central ways of replacing a module by a homotopically robust complex whose terms have good lifting or extension properties. [/explanation] The projective and injective constructions are not symmetric in appearance, but they serve the same structural purpose. Projective resolutions begin with surjections from projectives and measure kernels; injective resolutions begin with injections into injectives and measure cokernels. Both come with comparison maps and homotopy uniqueness, so both provide stable inputs for later constructions. [remark: What Has Not Yet Been Defined] This course has deliberately avoided defining the standard derived functors in detail. The names that appear in later courses are built from the machinery here: choose a suitable resolution, apply a functor term by term, and take homology or cohomology of the resulting complex. The hard work completed here is the proof that the answer does not depend on the chosen resolution. [/remark] The chapter also closes the loop with the exercises from earlier parts of the course. After separating the projective and injective constructions, we need a final synthesis theorem that names the common principle they support: resolutions let one replace a module by a complex whose terms are better suited to exactness arguments, while homotopy controls the dependence on choices. The snake lemma, connecting homomorphisms, mapping cones, projective modules as summands of free modules, and explicit projective resolutions are all examples of the same method: make the algebra visible by embedding it into a complex where exactness and homotopy can be checked degree by degree. [quotetheorem:4564] [citeproof:4564] This is the conceptual endpoint of the course, but the summary is only as strong as its hypotheses. Projective resolutions use the fact that modules are quotients of free modules, while injective resolutions use the enough-injectives property for module categories; outside those settings, one of the two constructions may fail. The finitely generated abelian groups again show the injective obstruction, since nonzero finitely generated groups cannot embed into injective objects of that category. The theorem also does not claim that projective and injective resolutions of the same module are homotopy equivalent to each other; they support different derived constructions such as $\operatorname{Ext}$, sheaf cohomology, and the later theory of derived functors. Complexes provide the language, exact sequences provide the tests, homotopy provides the equivalence relation, and resolutions provide the replacements that make later homological invariants computable.