Surjection iff Right Inverse/Proof 1

Proof
Assume $\exists g: T \to S: f \circ g = I_T$.

From Identity Mapping is Surjection, $I_T$ is surjective, so $f \circ g$ is surjective.

So from Surjection if Composite is Surjection, $f$ is a surjection.

Note that the existence of such a $g$ requires that $S$ is non-empty.

Now, assume $f$ is a surjection.

Consider the indexed family of non-empty sets $\set {\map {f^{-1} } y}_{y \mathop \in T}$ where $\map {f^{-1} } y$ denotes the preimage of $y$ under $f$.

Using the axiom of choice, there exists a mapping $g: T \to S$ such that $\map g y \in \set {\map {f^{-1} } y}$ for all $y \in T$.


 * SurjectionIffRightInverse.png

That is, $\map {\paren {f \circ g} } y = y$, as desired.