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 $\alpha_1(b)=\alpha_1(f(a))=(\alpha_1\circ f)(a)=(\alpha_2\circ f)(a)=\alpha_2(f(a))=\alpha_2(b)$.

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    $\begingroup$ I don't understand your question. In the category of sets and functions, the epimorphsms are the surjective maps. $\endgroup$ Mar 7, 2018 at 11:45
  • $\begingroup$ The question is not about the category of sets. I wrote that $f$ is a map of sets, because I want to talk about surjections. Suppose we are in the category of rings, then not every epimorphism is surjective. $\endgroup$
    – lanskey
    Mar 7, 2018 at 11:52
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    $\begingroup$ If you do not specify the category the question is about, then it is impossible to answer the question. $\endgroup$ Mar 7, 2018 at 12:02
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    $\begingroup$ The standard definition is: $f\colon A\to B$ is epi if for any two maps $\alpha_1,\alpha_2\colon B\to C$ $\alpha_1\circ f=\alpha_2\circ f\implies\alpha_1=\alpha_2$. $\endgroup$
    – lanskey
    Mar 7, 2018 at 12:41
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    $\begingroup$ A key point: if $f$ is surjective, you do not need the axiom of choice to say "for every $b$ we can find an $a$ with $f(a) = b$". That is just the definition of a surjection. The axiom of choice would be needed to generate a right inverse, that is, to form a function $g$ so that $f(g(b)) = b$ for $b$ in the r. But the proof above just works one element at a time, it does not need a right inverse of $f$. $\endgroup$ Mar 7, 2018 at 17:26

1 Answer 1


Even without choice surjections are epis in $\mathsf{Set}$:

Consider $e$ surjective and $f_1,f_2$ s.t. $f_1\circ e = f_2 \circ e$ and let $x$ be in the codomain of $e$. Then there exists $a$ s.t. $e(a) = x$, so $$f_1(x) = f_1(e(a)) = f_2(e(a)) = f_2(x).$$

Actually, I think your confusion comes from missusing the axiom of choice. In your proof (which is pretty much the same) it is simply not necessary. You use the axiom of choice for the other direction, i.e. surjections split.

For other concrete categories, the forgetful functor is by definition faithful and so it reflects monos and epis, hence surjective morphisms are epis.


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