[proofplan] We first reduce the proof in an arbitrary abelian category to an ordinary element chase by using the exact faithful embedding principle for the finite diagram involved: kernels, cokernels, exactness, and commutativity may be checked after embedding the relevant abelian subcategory into a module category. In that module category, we construct the connecting morphism by lifting a cycle in $C_i$ to an element of $B_i$, applying the differential in $B_\bullet$, and using exactness of $0 \to A_{i-1} \to B_{i-1} \to C_{i-1} \to 0$ to identify the result with a cycle in $A_{i-1}$. We then check that this assignment is independent of all choices and kills boundaries, so it descends to homology. The exactness and naturality obtained after the faithful exact embedding reflect back to the original abelian category. [/proofplan] [step:Reduce the abelian categorical proof to a finite element chase in a module category] The proof below uses elements only after a fully faithful exact embedding of the finite abelian diagram under discussion. Let $\mathcal{A}$ be the given abelian category, and let $\mathcal{D}$ be the finite diagram in $\mathcal{A}$ consisting of the objects and morphisms in the degrees needed to define and test \begin{align*} H_{i+1}(C_\bullet) \xrightarrow{\partial_{i+1}} H_i(A_\bullet) \to H_i(B_\bullet) \to H_i(C_\bullet) \xrightarrow{\partial_i} H_{i-1}(A_\bullet), \end{align*} together with the analogous finite diagram for a morphism of short exact sequences when naturality is considered. Let $\mathcal{A}_{\mathcal{D}}$ denote the essentially small full abelian subcategory generated by this finite diagram under finite direct sums, kernels, cokernels, images, and the homology objects appearing above. By the finite-diagram form of the Freyd-Mitchell embedding theorem, there are a ring $R$ and an exact fully faithful additive functor \begin{align*} F: \mathcal{A}_{\mathcal{D}} &\to R\operatorname{-Mod}. \end{align*} The hypotheses of this embedding result are satisfied because $\mathcal{A}_{\mathcal{D}}$ is an essentially small abelian category by construction. Exactness of $F$ means that it preserves kernels, cokernels, images, homology objects of chain complexes, and short exact sequences. Fullness means that every $R$-linear morphism between objects in the image of $F$ is the image of a morphism in $\mathcal{A}_{\mathcal{D}}$, and faithfulness means that equalities of such morphisms may be checked after applying $F$. Therefore it suffices to construct the connecting maps, verify exactness, and verify naturality after applying $F$; fullness lifts the constructed module homomorphisms back to morphisms in $\mathcal{A}_{\mathcal{D}}$, while exactness and faithfulness reflect the exactness and commutativity assertions. In the remaining steps we write the resulting module-category diagram with the same symbols; every occurrence of an element such as $c \in C_i$ is an element of the corresponding $R$-module after this exact fully faithful embedding. [guided] The theorem is stated in an arbitrary abelian category, so an ordinary element chase is not literally meaningful at the outset: an object of an abelian category need not have elements, and an epimorphism need not mean a surjective function on underlying sets. The way to make the classical proof rigorous is to reduce the finite diagram under discussion to a module category through an embedding that is exact, faithful, and full. Fix the degree $i$ being checked. The construction of $\partial_i$, the exactness at the adjacent homology objects, and the naturality square only involve finitely many objects and morphisms of the original chain complexes. Let $\mathcal{D}$ denote this finite diagram, and let $\mathcal{A}_{\mathcal{D}}$ be the full abelian subcategory generated by it under finite direct sums, kernels, cokernels, images, and the homology objects appearing in the argument. This is an essentially small abelian category: it is generated from a finite set of objects and morphisms by the finite categorical operations just listed. We now apply the finite-diagram form of the Freyd-Mitchell embedding theorem. Its hypotheses require an essentially small abelian category, which is precisely $\mathcal{A}_{\mathcal{D}}$. Hence there are a ring $R$ and an exact fully faithful additive functor \begin{align*} F: \mathcal{A}_{\mathcal{D}} &\to R\operatorname{-Mod}. \end{align*} Why is full faithfulness essential here? Exactness alone preserves kernels, cokernels, images, and short exact sequences, so the homology objects computed in $\mathcal{A}_{\mathcal{D}}$ become the ordinary homology modules after applying $F$. Faithfulness alone reflects equalities of morphisms. But the connecting map is first constructed by an element chase as an $R$-[linear map](/page/Linear%20Map) between homology modules; to know that this $R$-linear map actually comes from a morphism in the original abelian category, we need fullness. Fullness supplies a unique morphism in $\mathcal{A}_{\mathcal{D}}$ whose image under $F$ is the constructed connecting homomorphism. Consequently, the element chase below is not an additional hypothesis on $\mathcal{A}$. It constructs module homomorphisms after applying the exact fully faithful embedding. Fullness lifts those homomorphisms back to the original abelian category, and exactness plus faithfulness transfer the exactness and naturality verifications back to $\mathcal{A}$. [/guided] [/step] [step:Construct the connecting morphism from lifted cycles] Fix $i \in \mathbb Z$. After the reduction in the previous step, all objects are $R$-modules and all morphisms are $R$-linear maps. Write \begin{align*} Z_i(X_\bullet) &:= \ker(d_i^X : X_i \to X_{i-1}),& \operatorname{Bd}_i(X_\bullet) &:= \operatorname{im}(d_{i+1}^X : X_{i+1} \to X_i) \end{align*} for cycles and boundaries in a chain complex $X_\bullet$. Let $c \in Z_i(C_\bullet)$ be a cycle. Since $\beta_i : B_i \to C_i$ is an epimorphism, choose $b \in B_i$ with \begin{align*} \beta_i(b)=c. \end{align*} Because $\beta_\bullet$ is a chain map and $c$ is a cycle, \begin{align*} \beta_{i-1}(d_i^B b) = d_i^C(\beta_i b) = d_i^C c = 0. \end{align*} Exactness of \begin{align*} 0 \longrightarrow A_{i-1} \xrightarrow{\alpha_{i-1}} B_{i-1} \xrightarrow{\beta_{i-1}} C_{i-1} \longrightarrow 0 \end{align*} therefore gives a unique $a \in A_{i-1}$ such that \begin{align*} \alpha_{i-1}(a)=d_i^B b. \end{align*} This element $a$ is a cycle in $A_{i-1}$, because $\alpha_\bullet$ is a chain map and $(d^B)^2=0$: \begin{align*} \alpha_{i-2}(d_{i-1}^A a) = d_{i-1}^B(\alpha_{i-1} a) = d_{i-1}^B d_i^B b = 0. \end{align*} Since $\alpha_{i-2}$ is a monomorphism, $d_{i-1}^A a=0$. Define \begin{align*} \partial_i([c]) := [a] \in H_{i-1}(A_\bullet). \end{align*} [guided] Fix an integer $i$. By the previous step, we are now working after an exact faithful embedding into a module category, so elements, lifts, and quotients are ordinary module-theoretic notions. For a chain complex $X_\bullet$, we use the notation \begin{align*} Z_i(X_\bullet) &:= \ker(d_i^X : X_i \to X_{i-1}),& \operatorname{Bd}_i(X_\bullet) &:= \operatorname{im}(d_{i+1}^X : X_{i+1} \to X_i). \end{align*} Thus $H_i(X_\bullet)$ is the quotient of $Z_i(X_\bullet)$ by $\operatorname{Bd}_i(X_\bullet)$. We now define the connecting map. Start with a homology class in $H_i(C_\bullet)$ and choose a representative cycle $c \in Z_i(C_\bullet)$. Because $\beta_i : B_i \to C_i$ is an epimorphism, there is some $b \in B_i$ satisfying \begin{align*} \beta_i(b)=c. \end{align*} The obstruction to $b$ being a cycle is $d_i^B b \in B_{i-1}$. We show that this obstruction actually comes from $A_{i-1}$. Since $\beta_\bullet$ is a chain map and $d_i^C c=0$, we compute \begin{align*} \beta_{i-1}(d_i^B b) = d_i^C(\beta_i b) = d_i^C c = 0. \end{align*} Exactness at $B_{i-1}$ says that $\ker \beta_{i-1}=\operatorname{im}\alpha_{i-1}$. Hence there is a unique $a \in A_{i-1}$ such that \begin{align*} \alpha_{i-1}(a)=d_i^B b. \end{align*} The element $a$ is the candidate value of the connecting map. We must first check that $a$ is a cycle. Since $\alpha_\bullet$ is a chain map, \begin{align*} \alpha_{i-2}(d_{i-1}^A a) = d_{i-1}^B(\alpha_{i-1} a). \end{align*} Using the defining equation $\alpha_{i-1}(a)=d_i^B b$ and the chain complex identity $d_{i-1}^B d_i^B=0$, we obtain \begin{align*} \alpha_{i-2}(d_{i-1}^A a) = d_{i-1}^B d_i^B b = 0. \end{align*} Exactness at $A_{i-2}$ says that $\alpha_{i-2}$ is a monomorphism, so $d_{i-1}^A a=0$. Therefore $a \in Z_{i-1}(A_\bullet)$, and the formula \begin{align*} \partial_i([c]) := [a] \end{align*} at least produces a homology class in $H_{i-1}(A_\bullet)$. [/guided] [/step] [step:Show the connecting morphism is independent of choices] First suppose $b' \in B_i$ is another lift of the same cycle $c$, so $\beta_i(b'-b)=0$. Exactness at $B_i$ gives $u \in A_i$ such that \begin{align*} \alpha_i(u)=b'-b. \end{align*} If $a'$ is defined by $\alpha_{i-1}(a')=d_i^B b'$, then \begin{align*} \alpha_{i-1}(a'-a) = d_i^B(b'-b) = d_i^B\alpha_i(u) = \alpha_{i-1}d_i^A(u). \end{align*} Since $\alpha_{i-1}$ is a monomorphism, $a'-a=d_i^A(u)$, so $[a']=[a]$ in $H_{i-1}(A_\bullet)$. Next suppose $c$ is replaced by a homologous cycle $c+d_{i+1}^C e$, where $e \in C_{i+1}$. Choose $v \in B_{i+1}$ with $\beta_{i+1}(v)=e$. Then $b+d_{i+1}^B v$ is a lift of $c+d_{i+1}^C e$, and \begin{align*} d_i^B(b+d_{i+1}^B v) = d_i^B b+d_i^B d_{i+1}^B v = d_i^B b. \end{align*} Thus the resulting cycle in $A_{i-1}$ is the same $a$. It remains to check additivity. Let $c_1,c_2 \in Z_i(C_\bullet)$ be cycles, choose lifts $b_1,b_2 \in B_i$ with $\beta_i(b_j)=c_j$ for $j\in\{1,2\}$, and choose $a_1,a_2 \in A_{i-1}$ with \begin{align*} \alpha_{i-1}(a_j)=d_i^B b_j \end{align*} for $j\in\{1,2\}$. Then $b_1+b_2$ is a lift of $c_1+c_2$, and \begin{align*} \alpha_{i-1}(a_1+a_2) = d_i^B b_1+d_i^B b_2 = d_i^B(b_1+b_2). \end{align*} Hence $\partial_i([c_1]+[c_2])=[a_1+a_2]=[a_1]+[a_2]$. Similarly, for $r\in R$, the lift $rb_1$ of $rc_1$ gives the element $ra_1$, so $\partial_i(r[c_1])=r\partial_i([c_1])$. Therefore $\partial_i$ is a well-defined $R$-linear morphism \begin{align*} \partial_i : H_i(C_\bullet) \to H_{i-1}(A_\bullet). \end{align*} [guided] There are two possible choices in the construction: the lift $b$ of the cycle $c$, and the representative $c$ of the homology class $[c]$. First fix the representative $c$ and change only the lift. Let $b' \in B_i$ be another lift of $c$. Then \begin{align*} \beta_i(b'-b)=\beta_i(b')-\beta_i(b)=c-c=0. \end{align*} Exactness at $B_i$ gives an element $u \in A_i$ such that \begin{align*} \alpha_i(u)=b'-b. \end{align*} Let $a,a' \in A_{i-1}$ be the cycles produced from the lifts $b,b'$, so \begin{align*} \alpha_{i-1}(a)=d_i^B b, \qquad \alpha_{i-1}(a')=d_i^B b'. \end{align*} Subtracting and using that $\alpha_\bullet$ is a chain map, \begin{align*} \alpha_{i-1}(a'-a) = d_i^B(b'-b) = d_i^B\alpha_i(u) = \alpha_{i-1}d_i^A(u). \end{align*} Since $\alpha_{i-1}$ is a monomorphism, we may cancel it and obtain \begin{align*} a'-a=d_i^A(u). \end{align*} Thus $a$ and $a'$ differ by a boundary in $A_\bullet$, so they determine the same class in $H_{i-1}(A_\bullet)$. Now change the representative of the homology class. Suppose the new representative is $c+d_{i+1}^C e$ for some $e \in C_{i+1}$. Since $\beta_{i+1}$ is an epimorphism, choose $v \in B_{i+1}$ with $\beta_{i+1}(v)=e$. Then $b+d_{i+1}^B v$ lifts the new representative, because \begin{align*} \beta_i(b+d_{i+1}^B v) = \beta_i(b)+\beta_i d_{i+1}^B v = c+d_{i+1}^C\beta_{i+1}(v) = c+d_{i+1}^C e. \end{align*} Applying the differential gives \begin{align*} d_i^B(b+d_{i+1}^B v) = d_i^B b+d_i^B d_{i+1}^B v = d_i^B b, \end{align*} because $d_i^B d_{i+1}^B=0$. Hence the associated element of $A_{i-1}$ is exactly the same $a$. Finally, we check that the well-defined assignment is a module homomorphism. Let $c_1,c_2 \in Z_i(C_\bullet)$, and for $j\in\{1,2\}$ choose $b_j\in B_i$ and $a_j\in A_{i-1}$ satisfying \begin{align*} \beta_i(b_j)=c_j, \qquad \alpha_{i-1}(a_j)=d_i^B b_j. \end{align*} Then $b_1+b_2$ lifts $c_1+c_2$, and the differential calculation \begin{align*} \alpha_{i-1}(a_1+a_2) = d_i^B b_1+d_i^B b_2 = d_i^B(b_1+b_2) \end{align*} shows that $\partial_i([c_1]+[c_2])=[a_1+a_2]=[a_1]+[a_2]$. If $r\in R$, then $rb_1$ lifts $rc_1$, and \begin{align*} \alpha_{i-1}(ra_1)=r\alpha_{i-1}(a_1)=d_i^B(rb_1), \end{align*} so $\partial_i(r[c_1])=r\partial_i([c_1])$. Therefore the construction depends only on the homology class $[c]$ and defines a well-defined $R$-linear morphism \begin{align*} \partial_i : H_i(C_\bullet) \to H_{i-1}(A_\bullet). \end{align*} [/guided] [/step] [step:Prove exactness at $H_i(A_\bullet)$ and $H_i(B_\bullet)$] Exactness at $H_i(A_\bullet)$ means \begin{align*} \operatorname{im}(\partial_{i+1})=\ker H_i(\alpha_\bullet). \end{align*} If $[c]\in H_{i+1}(C_\bullet)$ and $\partial_{i+1}([c])=[a]$, then $\alpha_i(a)=d_{i+1}^B b$ for some lift $b\in B_{i+1}$ of $c$, so $H_i(\alpha_\bullet)([a])=0$. Conversely, if $[a]\in H_i(A_\bullet)$ and $H_i(\alpha_\bullet)([a])=0$, then $\alpha_i(a)=d_{i+1}^B b$ for some $b\in B_{i+1}$. Since \begin{align*} d_{i+1}^C\beta_{i+1}(b) = \beta_i d_{i+1}^B b = \beta_i\alpha_i(a) = 0, \end{align*} the element $\beta_{i+1}(b)$ is a cycle in $C_{i+1}$, and by construction \begin{align*} \partial_{i+1}([\beta_{i+1}(b)])=[a]. \end{align*} Exactness at $H_i(B_\bullet)$ means \begin{align*} \operatorname{im}H_i(\alpha_\bullet)=\ker H_i(\beta_\bullet). \end{align*} The inclusion $\operatorname{im}H_i(\alpha_\bullet)\subseteq\ker H_i(\beta_\bullet)$ follows from $\beta_i\alpha_i=0$. Conversely, if $[b]\in H_i(B_\bullet)$ and $H_i(\beta_\bullet)([b])=0$, then $\beta_i(b)=d_{i+1}^C c$ for some $c\in C_{i+1}$. Choose $v\in B_{i+1}$ with $\beta_{i+1}(v)=c$. Then \begin{align*} \beta_i(b-d_{i+1}^B v)=0, \end{align*} so exactness at $B_i$ gives $a\in A_i$ with $\alpha_i(a)=b-d_{i+1}^B v$. Since $b$ is a cycle, \begin{align*} \alpha_{i-1}d_i^A(a) = d_i^B\alpha_i(a) = d_i^B b-d_i^B d_{i+1}^B v = 0. \end{align*} Thus $a$ is a cycle, and $[b]=H_i(\alpha_\bullet)([a])$. [guided] We first prove exactness at $H_i(A_\bullet)$. We must show that the classes coming from $\partial_{i+1}$ are exactly the classes killed by $H_i(\alpha_\bullet)$. Let $[c]\in H_{i+1}(C_\bullet)$ and suppose $\partial_{i+1}([c])=[a]$. By definition of $\partial_{i+1}$, there is a lift $b\in B_{i+1}$ of $c$ such that \begin{align*} \alpha_i(a)=d_{i+1}^B b. \end{align*} The right-hand side is a boundary in $B_\bullet$, so the homology class of $\alpha_i(a)$ is zero: \begin{align*} H_i(\alpha_\bullet)([a])=[\alpha_i(a)]=0. \end{align*} Thus $\operatorname{im}(\partial_{i+1})\subseteq\ker H_i(\alpha_\bullet)$. Conversely, let $[a]\in H_i(A_\bullet)$ satisfy $H_i(\alpha_\bullet)([a])=0$. This means that $\alpha_i(a)$ is a boundary in $B_\bullet$, so there is $b\in B_{i+1}$ with \begin{align*} \alpha_i(a)=d_{i+1}^B b. \end{align*} Now apply $\beta_i$ to both sides. Since $\beta_i\alpha_i=0$ and $\beta_\bullet$ is a chain map, \begin{align*} d_{i+1}^C\beta_{i+1}(b) = \beta_i d_{i+1}^B b = \beta_i\alpha_i(a) = 0. \end{align*} Hence $\beta_{i+1}(b)$ is a cycle in $C_{i+1}$. The connecting construction applied to this cycle uses $b$ as a lift, and it sends $[\beta_{i+1}(b)]$ to $[a]$. Therefore \begin{align*} \ker H_i(\alpha_\bullet)\subseteq\operatorname{im}(\partial_{i+1}). \end{align*} Next prove exactness at $H_i(B_\bullet)$. The inclusion \begin{align*} \operatorname{im}H_i(\alpha_\bullet)\subseteq\ker H_i(\beta_\bullet) \end{align*} follows immediately from the degreewise exactness relation $\beta_i\alpha_i=0$. For the reverse inclusion, take a cycle $b\in Z_i(B_\bullet)$ whose class satisfies $H_i(\beta_\bullet)([b])=0$. This says that $\beta_i(b)$ is a boundary in $C_\bullet$, so there exists $c\in C_{i+1}$ with \begin{align*} \beta_i(b)=d_{i+1}^C c. \end{align*} Choose $v\in B_{i+1}$ with $\beta_{i+1}(v)=c$. Then \begin{align*} \beta_i(b-d_{i+1}^B v) = \beta_i(b)-d_{i+1}^C\beta_{i+1}(v) = d_{i+1}^C c-d_{i+1}^C c = 0. \end{align*} Exactness at $B_i$ gives $a\in A_i$ such that \begin{align*} \alpha_i(a)=b-d_{i+1}^B v. \end{align*} We check that $a$ is a cycle. Since $b$ is a cycle and $d_i^B d_{i+1}^B=0$, \begin{align*} \alpha_{i-1}d_i^A(a) = d_i^B\alpha_i(a) = d_i^B b-d_i^B d_{i+1}^B v = 0. \end{align*} Because $\alpha_{i-1}$ is a monomorphism, $d_i^A(a)=0$. Therefore $[a]\in H_i(A_\bullet)$, and the equation \begin{align*} \alpha_i(a)=b-d_{i+1}^B v \end{align*} shows that $[\alpha_i(a)]=[b]$ in $H_i(B_\bullet)$. Hence $[b]$ lies in the image of $H_i(\alpha_\bullet)$. [/guided] [/step] [step:Prove exactness at $H_i(C_\bullet)$] Exactness at $H_i(C_\bullet)$ means \begin{align*} \operatorname{im}H_i(\beta_\bullet)=\ker\partial_i. \end{align*} If $[b]\in H_i(B_\bullet)$, then $b$ itself is a cycle lifting $\beta_i(b)$, and the connecting construction gives $d_i^B b=0$, hence \begin{align*} \partial_i(H_i(\beta_\bullet)([b]))=0. \end{align*} Conversely, suppose $[c]\in H_i(C_\bullet)$ and $\partial_i([c])=0$. Choose $b\in B_i$ with $\beta_i(b)=c$, and let $a\in A_{i-1}$ satisfy $\alpha_{i-1}(a)=d_i^B b$. The equality $\partial_i([c])=0$ means that $a=d_i^A(u)$ for some $u\in A_i$. Then \begin{align*} d_i^B(b-\alpha_i(u)) = d_i^B b-\alpha_{i-1}d_i^A(u) = \alpha_{i-1}(a)-\alpha_{i-1}(a) = 0. \end{align*} Thus $b-\alpha_i(u)$ is a cycle in $B_i$, and \begin{align*} \beta_i(b-\alpha_i(u))=\beta_i(b)=c. \end{align*} Therefore $[c]=H_i(\beta_\bullet)([b-\alpha_i(u)])$, proving exactness at $H_i(C_\bullet)$. [guided] We must show that the image of $H_i(\beta_\bullet)$ is exactly the kernel of the connecting morphism $\partial_i$. First take a class $[b]\in H_i(B_\bullet)$. Since $b$ is a cycle, $d_i^B b=0$. Its image in $H_i(C_\bullet)$ is represented by $\beta_i(b)$. When constructing $\partial_i([\beta_i(b)])$, we may choose $b$ itself as the lift of $\beta_i(b)$. The differential of this lift is zero: \begin{align*} d_i^B b=0. \end{align*} Hence the corresponding element of $A_{i-1}$ is zero, and \begin{align*} \partial_i(H_i(\beta_\bullet)([b]))=0. \end{align*} Thus $\operatorname{im}H_i(\beta_\bullet)\subseteq\ker\partial_i$. Conversely, suppose $[c]\in H_i(C_\bullet)$ satisfies $\partial_i([c])=0$. Choose a lift $b\in B_i$ with $\beta_i(b)=c$. By the definition of $\partial_i$, there is $a\in A_{i-1}$ such that \begin{align*} \alpha_{i-1}(a)=d_i^B b, \end{align*} and $\partial_i([c])=[a]$. The assumption $\partial_i([c])=0$ means that $a$ is a boundary in $A_\bullet$, so there exists $u\in A_i$ with \begin{align*} a=d_i^A(u). \end{align*} Now modify the lift $b$ by subtracting $\alpha_i(u)$. This is designed to remove precisely the obstruction $d_i^B b$. Indeed, \begin{align*} d_i^B(b-\alpha_i(u)) = d_i^B b-d_i^B\alpha_i(u) = \alpha_{i-1}(a)-\alpha_{i-1}d_i^A(u) = 0. \end{align*} So $b-\alpha_i(u)$ is a cycle in $B_i$. It still maps to $c$, because $\beta_i\alpha_i=0$: \begin{align*} \beta_i(b-\alpha_i(u))=\beta_i(b)=c. \end{align*} Therefore $[c]$ is the image under $H_i(\beta_\bullet)$ of the class $[b-\alpha_i(u)]\in H_i(B_\bullet)$. This proves \begin{align*} \ker\partial_i\subseteq\operatorname{im}H_i(\beta_\bullet). \end{align*} [/guided] [/step] [step:Verify naturality of the long exact sequence] Let \begin{align*} \begin{array}{ccccccccc} 0 &\to& A_\bullet &\xrightarrow{\alpha_\bullet}& B_\bullet &\xrightarrow{\beta_\bullet}& C_\bullet &\to& 0\\ && \downarrow f_\bullet && \downarrow g_\bullet && \downarrow h_\bullet \\ 0 &\to& A'_\bullet &\xrightarrow{\alpha'_\bullet}& B'_\bullet &\xrightarrow{\beta'_\bullet}& C'_\bullet &\to& 0 \end{array} \end{align*} be a morphism of short exact sequences of chain complexes. Let $[c]\in H_i(C_\bullet)$, choose $b\in B_i$ with $\beta_i(b)=c$, and choose $a\in A_{i-1}$ with $\alpha_{i-1}(a)=d_i^B b$. Then $g_i(b)$ is a lift of $h_i(c)$, since \begin{align*} \beta'_i g_i(b)=h_i\beta_i(b)=h_i(c). \end{align*} Moreover, \begin{align*} \alpha'_{i-1} f_{i-1}(a) = g_{i-1}\alpha_{i-1}(a) = g_{i-1}d_i^B b = d_i^{B'}g_i(b). \end{align*} Thus the connecting construction for the lower short exact sequence sends $[h_i(c)]$ to $[f_{i-1}(a)]$. Hence \begin{align*} \partial'_i \circ H_i(h_\bullet) = H_{i-1}(f_\bullet)\circ \partial_i. \end{align*} The squares involving the maps induced by $\alpha_\bullet$ and $\beta_\bullet$ commute because the diagram of chain maps commutes degreewise. Therefore the entire long exact sequence is natural. [guided] Naturality means that a morphism between two short exact sequences of chain complexes induces a morphism between the corresponding long exact homology sequences. Let \begin{align*} \begin{array}{ccccccccc} 0 &\to& A_\bullet &\xrightarrow{\alpha_\bullet}& B_\bullet &\xrightarrow{\beta_\bullet}& C_\bullet &\to& 0\\ && \downarrow f_\bullet && \downarrow g_\bullet && \downarrow h_\bullet \\ 0 &\to& A'_\bullet &\xrightarrow{\alpha'_\bullet}& B'_\bullet &\xrightarrow{\beta'_\bullet}& C'_\bullet &\to& 0 \end{array} \end{align*} be such a morphism. This means $f_\bullet$, $g_\bullet$, and $h_\bullet$ are chain maps, and the diagram commutes in every degree: \begin{align*} g_i\alpha_i=\alpha'_i f_i, \qquad h_i\beta_i=\beta'_i g_i. \end{align*} We only need to check the square involving the connecting morphisms, because the squares involving $H_i(\alpha_\bullet)$ and $H_i(\beta_\bullet)$ follow directly from degreewise commutativity. Take $[c]\in H_i(C_\bullet)$. Choose $b\in B_i$ with $\beta_i(b)=c$, and choose $a\in A_{i-1}$ with \begin{align*} \alpha_{i-1}(a)=d_i^B b. \end{align*} Thus $\partial_i([c])=[a]$. Now apply the vertical maps. The element $g_i(b)\in B'_i$ lifts $h_i(c)\in C'_i$, because \begin{align*} \beta'_i g_i(b)=h_i\beta_i(b)=h_i(c). \end{align*} The differential of this lifted element is \begin{align*} d_i^{B'}g_i(b)=g_{i-1}d_i^B b, \end{align*} since $g_\bullet$ is a chain map. Using the defining equation for $a$ and the commutativity of the left square, we get \begin{align*} d_i^{B'}g_i(b) = g_{i-1}\alpha_{i-1}(a) = \alpha'_{i-1} f_{i-1}(a). \end{align*} Therefore the connecting construction in the lower short exact sequence sends $[h_i(c)]$ to $[f_{i-1}(a)]$. In symbols, \begin{align*} \partial'_i(H_i(h_\bullet)([c])) = H_{i-1}(f_\bullet)(\partial_i([c])). \end{align*} Since this holds for every homology class $[c]\in H_i(C_\bullet)$, \begin{align*} \partial'_i \circ H_i(h_\bullet) = H_{i-1}(f_\bullet)\circ \partial_i. \end{align*} Together with the already noted commutativity of the $\alpha$- and $\beta$-squares, this proves naturality of the long exact homology sequence. [/guided] [/step] [step:Assemble the long exact sequence] For each $i\in\mathbb Z$, the preceding construction gives a morphism \begin{align*} \partial_i : H_i(C_\bullet)\to H_{i-1}(A_\bullet). \end{align*} The exactness checks prove that every consecutive pair in \begin{align*} \cdots \longrightarrow H_i(A_\bullet) \xrightarrow{H_i(\alpha_\bullet)} H_i(B_\bullet) \xrightarrow{H_i(\beta_\bullet)} H_i(C_\bullet) \xrightarrow{\partial_i} H_{i-1}(A_\bullet) \xrightarrow{H_{i-1}(\alpha_\bullet)} H_{i-1}(B_\bullet) \longrightarrow \cdots \end{align*} has image equal to kernel. The naturality verification proves that these sequences commute with morphisms of short exact sequences of chain complexes. Since the element chase was carried out after an exact fully faithful embedding of the finite diagram into a module category, fullness lifts the constructed connecting homomorphisms back to morphisms in the original abelian category, while exactness and faithfulness reflect exactness and naturality. This completes the proof. [/step]