In this entry, we show the statement that

(*) every surjection has a left inverse

Proof. Let $f:A\to B$ be a surjection. Then the set $C:=\lbrace f^{-1}(y)\mid y\in B\rbrace$ partitions $A$ By the axiom of choice, there is a function $g:C\to \bigcup C$ such that $g(f^{-1}(y))\in f^{-1}(y)$ for every $y\in B$ Since $\bigcup C=A$ $g$ is a function from $C$ to $A$ Define $h:B\to A$ by $h(y)=g(f^{-1}(y))$ Then $h(y)\in f^{-1}(y)$ and therefore $(f\circ h)(y)=f(h(y))=y$ implying that $f$ has a left inverse.

Remark. The function $h$ is easily seen to be an injection: if $h(y_1)=h(y_2)$ then $y_1 = f(h(y_1))=f(h(y_2)) = y_2$

Before proving this, let us remark that, in the collection $C$ of non-empty sets of the axiom of choice, there is no assumption that the sets in $C$ be pairwise disjoint. The statement

(**) given a set $C$ of pairwise disjoint non-empty sets, there is a choice function $f:C\to \bigcup C$

Proof. Obviously AC implies (**). Conversely, assume (**). Let $C$ be a collection of non-empty sets. We assume $C\ne \varnothing$ For each $a\in C$ define a set $A_a:=\lbrace (x,a)\mid x\in a\rbrace$ Since $a\ne \varnothing$ $A_a\ne \varnothing$ In addition, $A_a\cap A_b= \varnothing$ iff $a\ne b$ (true since elements of $A_a$ and elements of $A_b$ have distinct second coordinates). So the collection $D:=\lbrace A_a\mid a\in C\rbrace$ is a set consisting of pairwise disjoint non-empty sets. By (**), there is a function $f:D\to \bigcup D$ such that $f(A_a)\in A_a$ for every $a\in C$ Now, define two functions $g: C\to D$ and $h:\bigcup D\to \bigcup C$ by $g(a)=A_a$ and $h(x,a)=x$ Then, for any $a\in C$ we have $(h\circ f \circ g)(a)=h(f(A_a))$ Since $f(A_a)\in A_a$ its first coordinate is an element of $a$ Therefore $h(f(A_a))\in a$ and hence $h\circ f\circ g$ is the desired choice function.

Remark. In the category of sets, AC is equivalent to saying that every epimorophism is a split epimorphism. In general, a category is said to have the axiom of choice if every epimorphism is a split epimorphism.