Injection has Surjective Left Inverse Mapping/Proof 1

Proof
Since $S$ is non-empty, we can choose an element $x \in S$.

Since $f$ is an injection, for each $t \in \Img f$ there exists a unique $s \in S$ such that:
 * $\map f s = t$

Thus by Law of Excluded Middle there exists a well-defined mapping $T \to S$ given by:
 * $\map g t = \begin {cases} s & : \paren {t \in \Img f} \land \paren {\map f s = t} \\ x & : t \notin \Img f \end {cases}$

By construction, for any given $s \in S$, the element $\map f s$ maps to $s$ under $g$.

Therefore $g: T \to S$ is a surjection.