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Suppose that $f: A \rightarrow B$ is epic in Sets. One can construct a section $s: B \rightarrow A$ of $f$ as follow: Let us define an equivalence relation $R$ on $A$ as follow: $aRa'$ iff $a, a' \in f^{-1}(b)$. This relation is clearly reflexive (because $f$ is surjective), symmetric, and transitive.

Thus, one can define the projection $\pi: A \rightarrow A/R$ and an injection $i: A/R \rightarrow A$ like $i([a]) = a$. Now, one can easily define a map $g: B \rightarrow A/R$ as $g(b) = f^{-1}(b)$.

Clearly, for $s = i \circ g$, one has $f \circ s = Id_B$.

I don't understand why axiom of choice is necessarily needed to define the injection map. Does this mean that the above construction fails without AC? Can someone explain me why and how is axiom of choice used to define $i$ ?

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    $\begingroup$ $i$ is not well-defined. $\endgroup$
    – Martin Brandenburg
    Mar 25, 2015 at 14:24

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The problem is that $i$ is not well-defined like that unless $f$ is an isomorphism. The axiom of choice guarantees that you can choose a representative of each equivalence class of $R$.

Let $A_b$ be the equivalence class corresponding to $b\in B$. Then $\prod_{b\in B} A_b\neq \emptyset$ by the axiom of choice. Thus, there exists $(a_b)_{b\in B}\in \prod_{b\in B}A_b$. Now define $s(b)=a_b$. It is easy to check that this is a section of $f$.

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    $\begingroup$ Ah you're right, I'm dumb, it has to be invariant by change of representative. $\endgroup$
    – sure
    Mar 25, 2015 at 14:25
  • $\begingroup$ How is the axiom of choice used formally there? I'm not sure to see how to use it explicitly to have a well defined injection. $\endgroup$
    – sure
    Mar 25, 2015 at 14:27
  • $\begingroup$ @sure Which formulation of the axiom of choice do you use? $\endgroup$
    – Julian Kuelshammer
    Mar 25, 2015 at 14:28
  • $\begingroup$ The one you wish :> but I would like to use the one about product of non empty sets is non empty $\endgroup$
    – sure
    Mar 25, 2015 at 14:28
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    $\begingroup$ Julian, if one wants to be accurate, then $(a_b)_{b\in B}=s$, because the elements of the product are exactly choice functions from the $A_b$'s. $\endgroup$
    – Asaf Karagila ♦
    Mar 25, 2015 at 14:43

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