injective map and split mono

Proposition

Let \(f: X \rightarrow Y\) be a map between sets where \(X\) is nonempty

\(f\) is injective
\(f\) is split mono, i.e. there exists a map \(g: Y \rightarrow X\) such that

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

Proof

1) \(\implies\) 2)

choose an element \(x_0 \in X\) (note that \(X\) is by assumption nonempty).
Then define

\begin{align*} g: Y \rightarrow& X \\ y \mapsto& \begin{cases} x_0 & \mbox{if } f^{-1}[ \{y\}] = \emptyset \\ x & \mbox{if } x \in f^{-1}[ \{y\}] \\ \end{cases} \end{align*}

which is welldefined since the fibre \(f^{-1}[ \{y\}]\) of an injective map is at most a singleton

Then this map \(g\) is a right inverse

2) \(\implies\) 1)

Let \(x_1,x_2 \in X\) such that

\begin{align*} f(x_1) = f(x_2) \end{align*}

Then we get

\begin{align*} g(f(x_1)) =& g(f(x_2)) \\ (g \circ f)(x_1) =& (g \circ f)(x_2) \\ \mathrm{id}(x_1) =& \mathrm{id}(x_2) \\ x_1 =& x_2 \end{align*}