= Image of a function
{title2=$f(X)$}
{wiki=Image_(mathematics)}
= Image
{synonym}
= Images of a function
{synonym}
For a <function> $f:X\to Y$ and a <subset> $E\subseteq X$, its image is
$$
f(E)=\{f(x):x\in E\}
=\{y\in Y:\text{ there exists }x\in E\text{ with }f(x)=y\}.
$$
The full <image of a function> is $f(X)$. The <set> notation ignores repeated outputs: in the finite example above, $f(\{a,b\})=\{0\}$, rather than a list containing zero twice. A <function collision> can therefore reduce the <cardinality> of an image.
<Images of a function> preserve <unions>:
$$
f(E\cup F)=f(E)\cup f(F).
$$
An output reached from the <union> is reached from at least one of its two input <sets>. For <intersections>, however, only
$$
f(E\cap F)\subseteq f(E)\cap f(F)
$$
holds in general. In the finite example, take $E=\{a\}$ and $F=\{b\}$. Their <intersection> is the <empty set>, so $f(E\cap F)=\varnothing$. But $f(E)=f(F)=\{0\}$, giving $f(E)\cap f(F)=\{0\}$. Distinct witnesses in the two input <sets> can produce the same output.
\Image[/image-preimage-intersection-asymmetry.png]
{title=An <image of a function> can merge disjoint inputs, while a <preimage> tests both output conditions on the same input}
For an <injective function>, an output shared by $f(E)$ and $f(F)$ must come from the same input. That input lies in $E\cap F$, so equality holds. Conversely, preserving <intersections> for every pair of input <subsets> forces <injectivity>: two distinct inputs with the same output would give the singleton counterexample above. The image and <preimage> laws are stated in https://www.jirka.org/ra/html/sec_basicset.html[Jiří Lebl, Basic Analysis I, Propositions 0.3.15–0.3.16]; the finite example makes the missing reverse inclusion visible.
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