Surjection iff Right Inverse/Proof 1

Theorem
A mapping $f: S \to T, S \ne \varnothing$ is a surjection iff:
 * $\exists g: T \to S: f \circ g = I_T$

where:
 * $g$ is a mapping
 * $I_T$ is the identity mapping on $T$.

That is, if $f$ has a right inverse.

Proof

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

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

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

Note that the existence of such a $g$ requires that $S \ne \varnothing$.


 * Now, assume $f$ is a surjection. Then:


 * $\forall y \in T: f^{-1} \left({\left\{{y}\right\}}\right) \ne \varnothing$

Let $f^{-1} \left({\left\{{y}\right\}}\right) = X_y = \left\{{x_{y_1}, x_{y_2}, \ldots}\right\}$

Using the Axiom of Choice, for each $y \in T$ we can choose any of the elements $x_{y_1}, x_{y_2}, \ldots$ to be identified as $x_y$, and thereby define:


 * $g: T \to S: g \left({y}\right) = x_y$


 * SurjectionIffRightInverse.png

Then we see that $f \circ g \left({y}\right) = f \left({x_y}\right) = y$

and thus $f \circ g = I_T$.