Image (mathematics)

If x is a member of X, then the image of x under f, denoted f(x),[1] is the value of f when applied to x. f(x) is alternatively known as the output of f for argument x.

The image of a subset A ⊆ X under f, denoted ${\displaystyle f[A]}$, is the subset of Y which can be defined using set-builder notation as follows:[2]

${\displaystyle f[A]=\{f(x)\mid x\in A\}}$

When there is no risk of confusion, ${\displaystyle f[A]}$ is simply written as ${\displaystyle f(A)}$. This convention is a common one; the intended meaning must be inferred from the context. This makes f[.] a function whose domain is the power set of X (the set of all subsets of X), and whose codomain is the power set of Y. See § Notation below for more.

The image of a function is the image of its entire domain, also known as the range of the function.[3]

If R is an arbitrary binary relation on X×Y, then the set { y∈Y | xRy for some x∈X } is called the image, or the range, of R. Dually, the set { x∈X | xRy for some y∈Y } is called the domain of R.

• ${\displaystyle f^{\rightarrow }:{\mathcal {P}}(X)\rightarrow {\mathcal {P}}(Y)}$ with ${\displaystyle f^{\rightarrow }(A)=\{f(a)\;|\;a\in A\}}$
• ${\displaystyle f^{\leftarrow }:{\mathcal {P}}(Y)\rightarrow {\mathcal {P}}(X)}$ with ${\displaystyle f^{\leftarrow }(B)=\{a\in X\;|\;f(a)\in B\}}$

• ${\displaystyle f_{\star }:{\mathcal {P}}(X)\rightarrow {\mathcal {P}}(Y)}$ instead of ${\displaystyle f^{\rightarrow }}$
• ${\displaystyle f^{\star }:{\mathcal {P}}(Y)\rightarrow {\mathcal {P}}(X)}$ instead of ${\displaystyle f^{\leftarrow }}$

• An alternative notation for f[A] used in mathematical logic and set theory is f "A.[7][8]
• Some texts refer to the image of f as the range of f, but this usage should be avoided because the word "range" is also commonly used to mean the codomain of f.

For every function ${\displaystyle f:X\rightarrow Y}$ and all subsets ${\displaystyle A\subseteq X}$ and ${\displaystyle B\subseteq Y}$, the following properties hold:

Image	Preimage
${\displaystyle f(X)\subseteq Y}$	${\displaystyle f^{-1}(Y)=X}$
${\displaystyle f(f^{-1}(Y))=f(X)}$	${\displaystyle f^{-1}(f(X))=X}$
${\displaystyle f(f^{-1}(B))\subseteq B}$ (equal if ${\displaystyle B\subseteq f(X)}$, e.g. ${\displaystyle f}$ is surjective)[9][10]	${\displaystyle f^{-1}(f(A))\supseteq A}$ (equal if ${\displaystyle f}$ is injective)[9][10]
${\displaystyle f(f^{-1}(B))=B\cap f(X)}$	${\displaystyle (f\vert _{A})^{-1}(B)=A\cap f^{-1}(B)}$
${\displaystyle f(f^{-1}(f(A)))=f(A)}$	${\displaystyle f^{-1}(f(f^{-1}(B)))=f^{-1}(B)}$
${\displaystyle f(A)=\varnothing \Leftrightarrow A=\varnothing }$	${\displaystyle f^{-1}(B)=\varnothing \Leftrightarrow B\subseteq Y\setminus f(X)}$
${\displaystyle f(A)\supseteq B\Leftrightarrow \exists C\subseteq A:f(C)=B}$	${\displaystyle f^{-1}(B)\supseteq A\Leftrightarrow f(A)\subseteq B}$
${\displaystyle f(A)\supseteq f(X\setminus A)\Leftrightarrow f(A)=f(X)}$	${\displaystyle f^{-1}(B)\supseteq f^{-1}(Y\setminus B)\Leftrightarrow f^{-1}(B)=X}$
${\displaystyle f(X\setminus A)\supseteq f(X)\setminus f(A)}$	${\displaystyle f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)}$[9]
${\displaystyle f(A\cup f^{-1}(B))\subseteq f(A)\cup B}$[11]	${\displaystyle f^{-1}(f(A)\cup B)\supseteq A\cup f^{-1}(B)}$[11]
${\displaystyle f(A\cap f^{-1}(B))=f(A)\cap B}$[11]	${\displaystyle f^{-1}(f(A)\cap B)\supseteq A\cap f^{-1}(B)}$[11]

Also:

• ${\displaystyle f(A)\cap B=\varnothing \Leftrightarrow A\cap f^{-1}(B)=\varnothing }$

For functions ${\displaystyle f:X\rightarrow Y}$ and ${\displaystyle g:Y\rightarrow Z}$ with subsets ${\displaystyle A\subseteq X}$ and ${\displaystyle C\subseteq Z}$, the following properties hold:

• ${\displaystyle (g\circ f)(A)=g(f(A))}$
• ${\displaystyle (g\circ f)^{-1}(C)=f^{-1}(g^{-1}(C))}$

For function ${\displaystyle f:X\rightarrow Y}$ and subsets ${\displaystyle A_{1},A_{2}\subseteq X}$ and ${\displaystyle B_{1},B_{2}\subseteq Y}$, the following properties hold:

Image	Preimage
${\displaystyle A_{1}\subseteq A_{2}\Rightarrow f(A_{1})\subseteq f(A_{2})}$	${\displaystyle B_{1}\subseteq B_{2}\Rightarrow f^{-1}(B_{1})\subseteq f^{-1}(B_{2})}$
${\displaystyle f(A_{1}\cup A_{2})=f(A_{1})\cup f(A_{2})}$[11][12]	${\displaystyle f^{-1}(B_{1}\cup B_{2})=f^{-1}(B_{1})\cup f^{-1}(B_{2})}$
${\displaystyle f(A_{1}\cap A_{2})\subseteq f(A_{1})\cap f(A_{2})}$[11][12] (equal if ${\displaystyle f}$ is injective[13])	${\displaystyle f^{-1}(B_{1}\cap B_{2})=f^{-1}(B_{1})\cap f^{-1}(B_{2})}$
${\displaystyle f(A_{1}\setminus A_{2})\supseteq f(A_{1})\setminus f(A_{2})}$[11] (equal if ${\displaystyle f}$ is injective[13])	${\displaystyle f^{-1}(B_{1}\setminus B_{2})=f^{-1}(B_{1})\setminus f^{-1}(B_{2})}$[11]
${\displaystyle f(A_{1}\triangle A_{2})\supseteq f(A_{1})\triangle f(A_{2})}$ (equal if ${\displaystyle f}$ is injective)	${\displaystyle f^{-1}(B_{1}\triangle B_{2})=f^{-1}(B_{1})\triangle f^{-1}(B_{2})}$

The results relating images and preimages to the (Boolean) algebra of intersection and union work for any collection of subsets, not just for pairs of subsets:

• ${\displaystyle f\left(\bigcup _{s\in S}A_{s}\right)=\bigcup _{s\in S}f(A_{s})}$
• ${\displaystyle f\left(\bigcap _{s\in S}A_{s}\right)\subseteq \bigcap _{s\in S}f(A_{s})}$
• ${\displaystyle f^{-1}\left(\bigcup _{s\in S}B_{s}\right)=\bigcup _{s\in S}f^{-1}(B_{s})}$
• ${\displaystyle f^{-1}\left(\bigcap _{s\in S}B_{s}\right)=\bigcap _{s\in S}f^{-1}(B_{s})}$

(Here, S can be infinite, even uncountably infinite.)

With respect to the algebra of subsets described above, the inverse image function is a lattice homomorphism, while the image function is only a semilattice homomorphism (i.e., it does not always preserve intersections).