If a map of sets $f:A\to B$ is surjective then it is an epimorphism. Is it possible to prove this without the Axiom of Choice? I know that in order to prove that surjective maps have right inverses we need the AC, but the former statement is weaker (epimorphisms need not have right inverses), and I guess it is reasonable to ask whether we can weaken our assumptions. My attempts so far have failed.
This question is not about the category of sets, but rather about any category in which we can talk about surjections, so, I suppose it applies to any concrete category.
Here is a sketch of a proof that uses the Axiom of Choice:
Suppose $f\colon A\to B$ is surjective. Then for any $b\in B$ we can find (Axiom of Choice) $a\in A$ such that $f(a)=b$. Now, take any two maps $\alpha_1,\alpha_2\colon B\to C$ to arbitrary set $C$, such that $\alpha_1\circ f=\alpha_2\circ f$ and pick an element $b\in B$. We have $\alpha_1(b)=\alpha_1(f(a))=(\alpha_1\circ f)(a)=(\alpha_2\circ f)(a)=\alpha_2(f(a))=\alpha_2(b)$.