What does the Axiom of Choice have to do with right inversibility?

Question

I have encountered an exercise that asks to prove that, these two statements are equivalent:

1. every surjective function has a right inverse.

2. Axiom of choice.

Definition: Given a function $f$, we say that $f$ has a right inverse if there exists $g$ such that $f \circ g$ is the identity function.

I must admit, I can't see the connection... Any help?

Thanks! Shir

Answer 1

Note that if $f\colon A\to B$ is onto $B$ (and we can always take $B=\operatorname{rng}(f)$ for that), then $f^{-1}(b)$ is a non-empty set. Using the axiom of choice it is easy to construct an inverse.

On the other hand, if you have a family of disjoint sets, $\{A_i\mid i\in I\}$ then there is a function from $\bigcup A_i$ onto $I$, such that its inverse is exactly a choice function.