The image records the values that a map actually attains. A function is declared with a codomain, but the codomain may contain many elements that the function never reaches; the image separates the declared target from the realized target. This distinction is basic in [Set](/page/Set) theory, [Function](/page/Function) theory, [Linear Map](/page/Linear%20Map) theory, and algebra, where the image of a structure-preserving map often becomes a subobject of the codomain. For example, a map $f: A \to B$ may have codomain $B$ even when all values of $f$ lie in a smaller subset $C \subsetneq B$. The image is the canonical way to name that smaller subset without changing the original map. It is the target actually seen by the domain. ## Definition The first question a function asks is not only where it is allowed to land, but where it lands after all inputs are used. The image answers this by collecting all output values produced by elements of the domain. [definition: Image of a Function] Let $A$ and $B$ be sets, and let $f: A \to B$ be a function. The image of $f$ is the subset of $B$ defined by \begin{align*} \operatorname{im} f = f(A) := \{f(a) : a \in A\}. \end{align*} [/definition] To use images in arguments about restricted domains, we also need a version that sends only a chosen subset forward. This refinement is essential whenever a proof tracks how a particular part of the source moves under a function. Many arguments need the image not only for the whole domain, but for a selected part of it. This lets a map transfer local or restricted information from a subset of the domain into the codomain. [definition: Image of a Subset] Let $A$ and $B$ be sets, let $f: A \to B$ be a function, and let $S \subset A$. The image of $S$ under $f$ is the subset of $B$ defined by \begin{align*} f(S) := \{f(s) : s \in S\}. \end{align*} [/definition] In algebra, this subset version becomes more powerful once the function preserves operations, because the forward image may inherit structure from the source. We need the following definition to isolate the linear case, where reached outputs become a vector subspace rather than only a subset. [definition: Image of a Linear Map] Let $U$ and $V$ be vector spaces over a field $k$, and let $T: U \to V$ be a linear map. The image of $T$ is \begin{align*} \operatorname{im} T := \{T(u) : u \in U\} \subset V. \end{align*} [/definition] For a general homomorphism, the important question is not only which elements are hit, but whether the hit elements still form a structure of the same kind inside the codomain. Naming this subset uniformly lets later statements compare it with kernels, quotients, and exactness without repeating separate language for groups, rings, modules, and vector spaces. [definition: Image of a Homomorphism] Let $X$ and $Y$ be algebraic structures of the same type, and let $\varphi: X \to Y$ be a homomorphism. The image of $\varphi$ is \begin{align*} \operatorname{im} \varphi := \{\varphi(x) : x \in X\} \subset Y. \end{align*} [/definition] The phrase "same type" means that $X$ and $Y$ might be groups, rings, modules over a fixed ring, vector spaces over a fixed field, or another category where homomorphisms have been defined. In each setting, the structural content of the image is supplied by the operations preserved by $\varphi$. ## Equivalent Characterisations The definition says that the image is a set of values. A useful way to recognize the same subset is by membership: an element of the codomain belongs to the image exactly when it is hit by some input. [quotetheorem:9481] Once membership in the image is expressed as solvability of an equation, the next question is whether every point of the codomain is solvable in this sense. We need a separate definition for functions whose image fills the whole codomain. [definition: Surjective Function] Let $A$ and $B$ be sets, and let $f: A \to B$ be a function. The function $f$ is surjective if \begin{align*} \operatorname{im} f = B. \end{align*} [/definition] To compare forward movement with backward movement under a function, we also need the companion operation that starts with a subset of the codomain and returns the inputs that land there. This next definition gives the backward operation that behaves differently from images and is often the more stable tool. [definition: Preimage of a Subset] Let $A$ and $B$ be sets, let $f: A \to B$ be a function, and let $T \subset B$. The preimage of $T$ under $f$ is \begin{align*} f^{-1}(T) := \{a \in A : f(a) \in T\}. \end{align*} [/definition] The image and preimage are not symmetric operations. Images preserve unions well, while preimages preserve unions, intersections, and complements more faithfully. ## Examples A polynomial map can have a codomain much larger than its image. This example shows why the codomain and image should not be identified. [example: Squaring Map on the Real Line] Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x)=x^2$. We show that \begin{align*} \operatorname{im} f = [0,\infty). \end{align*} First, if $z \in \operatorname{im} f$, then $z=f(x)$ for some $x \in \mathbb{R}$, so \begin{align*} z=x^2. \end{align*} Since $x^2 \geq 0$ for every real number $x$, it follows that $z \in [0,\infty)$. Hence $\operatorname{im} f \subset [0,\infty)$. Conversely, let $y \in [0,\infty)$. Then $y \geq 0$, so the real number $\sqrt{y}$ is defined and satisfies \begin{align*} (\sqrt{y})^2=y. \end{align*} Because $\sqrt{y} \in \mathbb{R}$ and $f(\sqrt{y})=(\sqrt{y})^2=y$, we have $y \in \operatorname{im} f$. Hence $[0,\infty) \subset \operatorname{im} f$, and therefore $\operatorname{im} f=[0,\infty)$. Thus $f$ is not surjective as a map $\mathbb{R} \to \mathbb{R}$, since $-1 \in \mathbb{R}$ but $-1 \notin \operatorname{im} f$. If instead the codomain is changed to $[0,\infty)$, the same formula $f(x)=x^2$ becomes surjective because every $y \in [0,\infty)$ is hit by the input $\sqrt{y}$. [/example] The example also illustrates a common warning: changing the codomain changes whether the map is surjective, but it does not change the formula defining the outputs. Linear maps make the image visible as a span. This is the bridge from the set-theoretic definition to the column space of a matrix. [example: Image of a Matrix Transformation] Let $T: \mathbb{R}^2 \to \mathbb{R}^3$ be defined by \begin{align*} T(x_1,x_2) = (x_1+x_2,2x_1+2x_2,x_1-x_2). \end{align*} For each $(x_1,x_2) \in \mathbb{R}^2$, scalar multiplication and addition in $\mathbb{R}^3$ give \begin{align*} x_1(1,2,1)=(x_1,2x_1,x_1). \end{align*} Also, \begin{align*} x_2(1,2,-1)=(x_2,2x_2,-x_2). \end{align*} Adding these two vectors coordinatewise, \begin{align*} x_1(1,2,1)+x_2(1,2,-1)=(x_1+x_2,2x_1+2x_2,x_1-x_2). \end{align*} Therefore \begin{align*} T(x_1,x_2)=x_1(1,2,1)+x_2(1,2,-1). \end{align*} This shows that every value of $T$ lies in $\operatorname{span}\{(1,2,1),(1,2,-1)\}$, so \begin{align*} \operatorname{im} T \subset \operatorname{span}\{(1,2,1),(1,2,-1)\}. \end{align*} Conversely, if $v$ is any vector in that span, then for some $a,b \in \mathbb{R}$, \begin{align*} v=a(1,2,1)+b(1,2,-1). \end{align*} Using the computation above with $x_1=a$ and $x_2=b$ gives \begin{align*} v=T(a,b). \end{align*} Since $(a,b)\in \mathbb{R}^2$, this means $v \in \operatorname{im}T$. Hence \begin{align*} \operatorname{span}\{(1,2,1),(1,2,-1)\} \subset \operatorname{im} T. \end{align*} The two inclusions give \begin{align*} \operatorname{im} T=\operatorname{span}\{(1,2,1),(1,2,-1)\}. \end{align*} To check that the two spanning vectors are linearly independent, suppose \begin{align*} \alpha(1,2,1)+\beta(1,2,-1)=(0,0,0). \end{align*} Expanding the left-hand side gives \begin{align*} (\alpha+\beta,2\alpha+2\beta,\alpha-\beta)=(0,0,0). \end{align*} Equality of coordinates gives $\alpha+\beta=0$ and $\alpha-\beta=0$. Adding these two equations gives $2\alpha=0$, so $\alpha=0$, and then $\alpha+\beta=0$ gives $\beta=0$. Thus the spanning set has two linearly independent vectors, so $\operatorname{im}T$ is a two-dimensional subspace of $\mathbb{R}^3$. [/example] Here the image is smaller than the codomain but still carries the full structure expected from a linear map: it is closed under addition and scalar multiplication. Group homomorphisms show the same phenomenon in algebraic language. The outputs of a homomorphism form a subgroup, and that subgroup is the part of the codomain controlled by the source. [example: Image of a Modular Homomorphism] In this example, $\mathbb{Z}/6\mathbb{Z}$ denotes the integers modulo $6$: two integers represent the same element when their difference is divisible by $6$. The notation $\overline{a}$ means the residue class of the integer $a$ modulo $6$, so $\overline{a}=\overline{b}$ exactly when $a\equiv b\pmod 6$. Thus expressions such as $\overline{2n}$ name the residue class of the integer $2n$, not a new kind of integer. Let $\varphi: \mathbb{Z} \to \mathbb{Z}/6\mathbb{Z}$ be defined by \begin{align*} \varphi(n)=\overline{2n}. \end{align*} It is a [group homomorphism](/page/Group%20Homomorphism) because for $m,n \in \mathbb{Z}$, \begin{align*} \varphi(m+n)=\overline{2(m+n)}=\overline{2m+2n}=\overline{2m}+\overline{2n}=\varphi(m)+\varphi(n). \end{align*} We compute its image. Every integer $n$ has one of the forms $n=3q$, $n=3q+1$, or $n=3q+2$ for some $q \in \mathbb{Z}$. In these three cases, \begin{align*} \varphi(3q)=\overline{2(3q)}=\overline{6q}=\overline{0}. \end{align*} \begin{align*} \varphi(3q+1)=\overline{2(3q+1)}=\overline{6q+2}=\overline{2}. \end{align*} \begin{align*} \varphi(3q+2)=\overline{2(3q+2)}=\overline{6q+4}=\overline{4}. \end{align*} Thus every value of $\varphi$ is one of $\overline{0}$, $\overline{2}$, or $\overline{4}$, so \begin{align*} \operatorname{im}\varphi \subset \{\overline{0},\overline{2},\overline{4}\}. \end{align*} Conversely, \begin{align*} \varphi(0)=\overline{0}, \qquad \varphi(1)=\overline{2}, \qquad \varphi(2)=\overline{4}. \end{align*} Hence \begin{align*} \operatorname{im}\varphi=\{\overline{0},\overline{2},\overline{4}\}. \end{align*} This image is a subgroup of $\mathbb{Z}/6\mathbb{Z}$ under addition: it contains the identity $\overline{0}$, and its sums stay inside the same set because \begin{align*} \overline{2}+\overline{2}=\overline{4}, \qquad \overline{2}+\overline{4}=\overline{6}=\overline{0}, \qquad \overline{4}+\overline{4}=\overline{8}=\overline{2}. \end{align*} It is not the whole group $\mathbb{Z}/6\mathbb{Z}$, since $\overline{1} \notin \operatorname{im}\varphi$: if $\overline{2n}=\overline{1}$, then $2n \equiv 1 \pmod{6}$, so $2n=1+6q$ for some $q \in \mathbb{Z}$, but the left side is even while the right side is odd. [/example] This computation exhibits the image as an obstruction to solving equations inside the codomain: an element lies outside the image exactly when the corresponding equation has no solution in the source. Images can behave badly with intersections. This is why many set identities involving images need hypotheses such as injectivity. [example: Image Does Not Preserve Intersections] Let $A=\{-1,1\}$, let $B=\{0,1\}$, and define $f: A \to B$ by \begin{align*} f(-1)=0, \qquad f(1)=0. \end{align*} Set $S=\{-1\}$ and $T=\{1\}$. Since $-1 \neq 1$, there is no element lying in both $S$ and $T$, so \begin{align*} S \cap T=\varnothing. \end{align*} Therefore \begin{align*} f(S \cap T)=f(\varnothing)=\{f(a):a\in \varnothing\}=\varnothing. \end{align*} On the other hand, the image of $S$ is \begin{align*} f(S)=\{f(a):a\in \{-1\}\}=\{f(-1)\}=\{0\}. \end{align*} Similarly, the image of $T$ is \begin{align*} f(T)=\{f(a):a\in \{1\}\}=\{f(1)\}=\{0\}. \end{align*} Hence \begin{align*} f(S)\cap f(T)=\{0\}\cap \{0\}=\{0\}. \end{align*} Thus $f(S\cap T)=\varnothing$ while $f(S)\cap f(T)=\{0\}$, so \begin{align*} f(S\cap T)\subsetneq f(S)\cap f(T). \end{align*} The failure occurs because the two distinct inputs $-1$ and $1$ are sent to the same output $0$. [/example] To use intersections safely, we need a theorem that records exactly what always remains true and what improves when the map is injective. This prevents the common mistake of treating images as though they behaved like preimages. ## Properties The most reliable set operation for images is union. A point lies in the image of a union exactly when it comes from at least one of the pieces. [quotetheorem:9482] This is why images are easy to use when an object is built by adjoining pieces. If a set is presented as a union of simpler subsets, its image can be computed piece by piece and then reassembled by taking the union of the individual images. The theorem also marks the main limitation of this convenience: it is a statement about existence of at least one preimage, so it does not control whether different pieces overlap after applying the map. Intersections require a separate statement because two distinct inputs can merge into the same output. Thus an output can lie in both $f(S)$ and $f(T)$ without coming from an element that lies in both $S$ and $T$. The useful result is the exact containment that always survives, together with the extra hypothesis under which equality is restored. [quotetheorem:9483] This result explains exactly what went wrong in the preceding counterexample. The inclusion $f(S\cap T)\subseteq f(S)\cap f(T)$ never needs injectivity: an element of $S\cap T$ certainly maps to something lying in both images. The reverse inclusion is the fragile direction, because a point of $f(S)\cap f(T)$ may have one preimage in $S$ and a different preimage in $T$. Injectivity is precisely the condition that prevents those two preimages from being different, so it converts the surviving containment into equality. After the set-theoretic identities are in place, the next key issue is what extra structure the image has when the map preserves algebraic operations. In the linear setting, the image is a subspace, so it can be studied by bases and dimension. There is also a second issue: a linear map may collapse different inputs to the same output, and the collapsed directions are exactly its [kernel](/page/Kernel), the subspace of inputs sent to the zero vector. For a subspace $W\subseteq U$, the quotient [vector space](/page/Vector%20Space) $U/W$ is the vector space whose elements are cosets $u+W=\{u+w:w\in W\}$. Two cosets $u+W$ and $u'+W$ are equal precisely when $u-u'\in W$. Thus, for a linear map $\alpha:U\to V$, the quotient $U/\ker\alpha$ identifies exactly those vectors that $\alpha$ cannot distinguish. The [First Isomorphism Theorem](/theorems/791) says that after making precisely this identification, the remaining information in the domain is exactly the image. [quotetheorem:384] The quotient in this theorem is necessary: without dividing by $\ker\alpha$, non-injective maps still remember artificial distinctions between inputs that have the same output. The theorem says that once those distinctions are collapsed, the remaining source-side information matches exactly the part of $V$ that is actually reached. In the displayed isomorphism, $U/\ker\alpha$ is the corrected source and $\operatorname{im}\alpha$ is the natural target, so the image appears as the structure that remains after the irrelevant kernel directions have been removed. The same structural inheritance is not limited to vector spaces, so the homomorphism version is the next algebraic principle. Before stating it, it is useful to introduce exactness, where images are compared directly with kernels. The image also appears in exactness. In an exact sequence, the outputs of one map are precisely the inputs killed by the next map. [definition: Exactness at an Object] Let \begin{align*} A \xrightarrow{f} B \xrightarrow{g} C \end{align*} be homomorphisms of groups, modules over a fixed ring, or vector spaces over a fixed field. The sequence is exact at $B$ if \begin{align*} \operatorname{im} f = \ker g. \end{align*} [/definition] Since many algebraic maps are homomorphisms, one needs a general structural fact saying that the image remains inside the codomain as a legitimate algebraic object. The next theorem is what allows images to appear in quotient theorems and exact sequences without leaving the category under discussion. [quotetheorem:9484] The conclusion is what makes formulas such as $\operatorname{im} f=\ker g$ meaningful inside algebra: both sides are subobjects of the same ambient object, not merely subsets. The homomorphism hypotheses are essential because closure under the relevant operations has to be inherited from the domain and transported through the map. Without operation preservation, an arbitrary set-theoretic image inside a group, ring, vector space, or module need not be closed under the operations required to be a subgroup, subring, subspace, or submodule. Once the image has been identified as the part of the codomain actually reached, every map has a built-in redundancy: its declared codomain may contain many elements that play no role. This raises a basic structural question about any homomorphism: can it be decomposed into a map that reaches all of its target, followed by the inclusion of that reached subobject into the original codomain? The following factorization records that universal way of separating the genuinely produced outputs from the ambient codomain in which they sit. [quotetheorem:9485] This factorization separates the two roles of a map: first reaching every point it actually reaches, then including those reached points into the declared codomain. The decomposition is useful because surjectivity onto the image is automatic, while the inclusion of the image into the codomain records exactly the ambient object in which those outputs are being viewed. It is also limited in an important way: it does not remove identifications among elements of the domain. For example, a non-injective homomorphism can still collapse many inputs before it reaches the image; the later quotient theorems handle that collapse by dividing out the kernel. Thus the canonical factorization isolates the codomain redundancy, while kernel quotients isolate the domain redundancy. ## Relationship to Other Concepts The image is paired with the [Kernel](/page/Kernel) in algebra. The kernel measures what collapses to the neutral or zero element, while the image measures what is reached. Together they control many classification results. For a group homomorphism $\varphi:G\to H$, the kernel $\ker(\varphi)$ is a [normal subgroup](/page/Normal%20Subgroup) of $G$, written $\ker(\varphi)\trianglelefteq G$, so the [quotient group](/theorems/790) $G/\ker(\varphi)$ is defined. Its elements are cosets $g\ker(\varphi)=\{gk:k\in\ker(\varphi)\}$. Two elements of $G$ determine the same coset exactly when they differ by an element killed by $\varphi$. With this notation, the group version of the First Isomorphism Theorem says that the image is the group obtained after making precisely those identifications. [quotetheorem:842] This theorem explains why the image is not just a collection of outputs: it is the source group after the kernel-based redundancies have been removed. The displayed isomorphism should be read as a structural identification between the quotient $G/\ker(\varphi)$ and the reached subgroup $\operatorname{im}(\varphi)$. The coset notation records the bookkeeping of that identification, while the conceptual point is that the image keeps exactly the group structure visible from the homomorphism. In linear algebra, the image is the range part of the kernel-image balance measured by rank and nullity. The theorem below is the dimension-level version of that balance. [quotetheorem:385] The [rank-nullity theorem](/theorems/916) turns the image into a numerical invariant. Its dimension is the rank of $T$, and it counts the independent directions in the codomain that the map can reach. In topology, images interact with [Continuity](/page/Continuity) in a different way. Continuous functions are defined through preimages of open sets, not images of open sets, precisely because images of open sets need not be open. [example: Continuous Image of an Open Set Need Not Be Open] Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x)=x^2$. First, $f$ is continuous: for a fixed $a\in\mathbb{R}$ and any $x\in\mathbb{R}$, \begin{align*} |f(x)-f(a)|=|x^2-a^2|=|x-a||x+a|. \end{align*} If $|x-a|<1$, then $|x+a|\leq |x-a|+2|a|<1+2|a|$. Given $\varepsilon>0$, choose \begin{align*} \delta=\min\left\{1,\frac{\varepsilon}{1+2|a|}\right\}. \end{align*} Then $|x-a|<\delta$ implies \begin{align*} |f(x)-f(a)|<\delta(1+2|a|)\leq \varepsilon. \end{align*} Thus $f$ is continuous at every $a\in\mathbb{R}$. We compute the image of the open interval $(-1,1)$. If $x\in(-1,1)$, then $-1<x<1$, so $0\leq x^2<1$. Hence \begin{align*} f((-1,1))\subset [0,1). \end{align*} Conversely, let $y\in[0,1)$. Then $0\leq y<1$, so $\sqrt{y}$ is a real number satisfying $0\leq \sqrt{y}<1$. Therefore $\sqrt{y}\in(-1,1)$ and \begin{align*} f(\sqrt{y})=(\sqrt{y})^2=y. \end{align*} So $y\in f((-1,1))$, and therefore \begin{align*} f((-1,1))=[0,1). \end{align*} The set $[0,1)$ is not open in $\mathbb{R}$: for every $r>0$, the point $-r/2$ satisfies $-r<-r/2<r$, so $-r/2\in(-r,r)$, but $-r/2\notin[0,1)$. Thus no open interval around $0$ lies inside $[0,1)$. The continuous map $x\mapsto x^2$ therefore sends the [open set](/page/Open%20Set) $(-1,1)$ to a set that is not open, so continuity does not preserve openness of images. [/example] This example is a useful guardrail: the image is a forward operation, but many structural properties are better controlled by backward operations such as preimage. In category-theoretic language, images are part of the general study of how morphisms factor. In concrete algebra, that abstract idea returns to familiar objects: subgroups, subrings, submodules, subspaces, and quotient objects. [remark: Image and Range] The terms image and range are often used synonymously. On Androma, $\operatorname{im} f$ is preferred for the image of a map, while range may appear in prose when discussing the values attained by a function. [/remark] The essential distinction remains simple: the codomain is part of the data used to declare a function, while the image is determined by the values the function takes. ## Beyond and Connections Images connect the elementary language of functions with several larger themes across Androma. In set theory, images and preimages explain how functions transport subsets, and the contrast between unions and intersections is a first example of how non-injectivity affects that transport. In linear algebra and group theory, the image records the part of the target actually reached by a homomorphism, while the kernel records what the map collapses; the First Isomorphism Theorems make these two pieces complementary. Exactness uses images as a measuring device: saying that a sequence is exact at an object means that the image of one map is exactly the kernel of the next. Topological examples add another layer, because the set-theoretic image of an open or [closed set](/page/Closed%20Set) need not remain open or closed unless the map has additional properties. Thus the image is not merely a definition attached to functions; it is the common language behind range, surjectivity, quotient constructions, exact sequences, and the behavior of maps in topology. ## References - Paul R. Halmos, *Naive Set Theory* (1960). - Serge Lang, *Algebra* (2002). - Sheldon Axler, *Linear Algebra Done Right* (2015). - Michael Artin, *Algebra* (2011). - [Function](/page/Function). - [Linear Map](/page/Linear%20Map). - [Kernel](/page/Kernel).