surjektiv und split epi in ZFC

Proposition

Assume the Axiom of choice.
Let \(X,Y\) be sets and \(f: X \rightarrow Y\) a map

\(f\) is surjective
\(f\) is split epi i.e. there exists a right inverse \(g: Y \rightarrow X\) such that

\begin{align*} f \circ g = \mathrm{id}_Y \end{align*}

Proof

1) \(\implies\) 2)

Note that for each \(y \in Y\) the fibre \(f^{-1}[ \{y\}]\) is nonempty
Hence we may choose for each \(y \in Y\) an element \(x_y \in f^{-1}[ \{y\}]\).
Then define

\begin{align*} g: Y \rightarrow& X \\ y \mapsto& x_y \end{align*}

This \(g\) then provides a right inverse

2) \(\implies\) 1)

Let \(y \in Y\).
Then \(g(y)\) is a preimage of \(y\) since

\begin{align*} f(g(y)) =& (f \circ g)(y) \\ =& \mathrm{id}(y) \\ =& y \end{align*}