I've been studying some lecture notes on abstract algebra, and they claim at one point that a proof that a function $f: A \to B$ is injective if and only if it has a left inverse requires choice. The proof of surjectivity if only if there exists a right inverse, surely does (there may exist $b \in B$ with more than one, in fact infinitely many, preimages in $A$), but I don't believe the left-inverse proof requires it. This is what I have in mind. I'm going to skip the left inverse implies injective proof, as it surely doesn't require choice.
Let $f: A \to B$ be injective. Given $b \in B$, we have iether $b \in f(A)$ or $b \not \in f(A)$. If $b \in f(A)$, there exists a unique, since $f$ is injective, $a \in A$ such that $f(a) = b$. Fix $a_0 \in A$. Then define the map $g: B \to A$ sending $b \to a$ if $b = f(a)$ for some $a$ and $b \to a_0$ otherwise. Then for any $a \in A$, we have $(g \circ f)(a) = g(f(a)) = a$, so $g$ is a left inverse.
I don't believe I used the axiom of choice anywhere in this proof. Is that correct?
