= Preimage
{title2=$f^{-1}(B)=\{x:f(x)\in B\}$}
= Inverse image
{synonym}
For a <function> $f:X\to Y$ and a <subset> $U\subseteq Y$, its <preimage> is
$$
f^{-1}(U)=\{x\in X:f(x)\in U\}.
$$
It is a <subset> of the original <domain of a function>. The notation does not require an <inverse function>: the <preimage> of a <singleton set> can have no members, one member, or many members. In the finite example, $f^{-1}(\{0\})=\{a,b\}$, $f^{-1}(\{1\})=\{c\}$ and $f^{-1}(\{3\})=\varnothing$. The <preimage> of a <singleton set> is also its <fiber of a function>.
For <subsets> $U,V\subseteq Y$, <preimages> preserve both <unions> and <intersections>:
$$
f^{-1}(U\cup V)=f^{-1}(U)\cup f^{-1}(V),\qquad
f^{-1}(U\cap V)=f^{-1}(U)\cap f^{-1}(V).
$$
For the <intersection>, the same input $x$ belongs precisely when its single output $f(x)$ belongs to both $U$ and $V$. There is no opportunity to choose different witnesses, which is what made the image counterexample fail. The identities also hold for arbitrary families of output <subsets>.
In the finite example, choose $U=\{0,1\}$ and $V=\{0,2\}$. Then
$$
f^{-1}(U)=\{a,b,c\},\qquad f^{-1}(V)=\{a,b,d\},\qquad
f^{-1}(U\cap V)=f^{-1}(\{0\})=\{a,b\}.
$$
The <intersection> of the two preimages is exactly the same <set> $\{a,b\}$. Complements are preserved with respect to the appropriate source and target <sets>:
$$
f^{-1}(Y\setminus U)=X\setminus f^{-1}(U).
$$
For example, the unused output three contributes no extra input: with $U=\{0,2\}$, both sides of the complement identity are $\{c\}$. See https://www.jirka.org/ra/html/sec_basicset.html[Lebl, Proposition 0.3.15: preimages of unions, intersections and complements].
This makes <preimages> useful for transporting conditions. The <solution set> of $g(x)=0$ is $g^{-1}(\{0\})$, and a <zero set> is a preimage of zero. Requiring two output conditions takes the <intersection> of their preimages. In <topology>, a <continuous function> pulls every <open set> back to an <open set>; a <measurable function> pulls the specified <measurable sets> in its <codomain> back to <measurable sets> in its <domain of a function>. These definitions use <preimages> because their set operations remain consistent even when the <function> is noninjective.
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