Preimage

EVERYTHING THAT LANDS IN A CHOSEN SET

Name any subset of the target, and the preimage hands back every point of the domain that lands in it. It asks nothing of the map in return: points may pile up on one target point, target points may be missed entirely, and the answer is still a subset of the domain.

M N f 1 2 3 4 1 2 3 nothing lands here preimf({1}) = {1,2} two points of M land in it preimf({2}) = ∅ nothing lands in it at all preimf({1,3}) = {1,2,3,4} everything lands in it
One map on four points and three. Points can pile up and points can be missed; the preimage minds neither.

The definition

Definition · preimage

Let \(f : M \to N\) be a map, \(M\) the domain and \(N\) the target, and let \(V \subseteq N\). The preimage of \(V\) under \(f\) is:

\[ \mathrm{preim}_f(V) := \{\, p \in M \mid f(p) \in V \,\}. \]

Every point of \(M\) gets a verdict, since \(f\) gives it exactly one image to test. \(V\) may be any subset of \(N\) at all.

It is not an inverse

An inverse sends each point of \(N\) back to the one point it came from, and exists only when \(f\) is a bijection, injective and surjective at once. The preimage never needs that. Two points landing together give \(\mathrm{preim}_f(\{1\}) = \{1,2\}\), where an inverse would have to choose between them; a point nothing reaches gives \(\mathrm{preim}_f(\{2\}) = \emptyset\), where an inverse would have nothing to return. Both are ordinary answers.

It carries unions, intersections and complements across

Because a point has one image, the preimage carries the set operations across intact. For \(A, B \subseteq N\):

\[ \mathrm{preim}_f(A \cup B) = \mathrm{preim}_f(A) \cup \mathrm{preim}_f(B) \]\[ \mathrm{preim}_f(A \cap B) = \mathrm{preim}_f(A) \cap \mathrm{preim}_f(B) \]

and likewise for the complement, \(\mathrm{preim}_f(N \setminus A) = M \setminus \mathrm{preim}_f(A)\). Sending points forward keeps only the union: take two disjoint subsets of \(M\) that share an image point, and the image of their intersection is empty while the intersection of their images is not.