Exercise A0.V.3.03

Proposition

The axiom of choice is equivalent to the statement “every surjective set function has a right inverse.”

Proof

Suppose every surjective function has a right inverse. Let $\mathscr{F}$ be a family of disjoint nonempty subsets of $Z.$ Let $Z' = \bigcup \mathscr{F}.$ Let $g : Z' \to \mathscr{F}$ be the function sending each $z \in Z'$ to the unique set $A \in \mathscr{F}$ such that $z \in A.$ This function is surjective, since each $z \in Z'$ comes from an element of $\mathscr{F}.$

Let $g^{-1} : \mathscr{F} \to Z'$ be a right inverse to $g$. If $A \in \mathscr{F}$ then $a = g^{-1}(A)$ must satisfy $g(a) = A,$ and hence $a \in A.$ Then the image of $g^{-1}$ is exactly the set containing one element of each $A \in \mathscr{F},$ as required.

Conversely, suppose that the axiom of choice holds and let $f : A \to B$ be a surjective function. For each element $b \in B,$ the fiber $f^{-1}\{b\}$ is non-empty (by surjectivity) and any two distinct elements of $B$ have disjoint fibers (since $f$ is a function). A right inverse of $f$ can be constructed by choosing a single representative from each fiber using the axiom of choice.

$\blacksquare$