 Preimage

For a function $f:X\to Y$ and a subset $U\subseteq Y$, its preimage isIt 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: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 $V=\{0,2\}$. ThenThe intersection of the two preimages is exactly the same set $\{a,b\}$. Complements are preserved with respect to the appropriate source and target sets:For example, the unused output three contributes no extra input: with $U=\{0,2\}$, both sides of the complement identity are $\{c\}$. See 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.