Surjection iff Right Inverse

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.

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

In general, that right inverse is not unique.

Uniqueness occurs iff $f$ is an injection.

Proof 1

 * 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 $f \circ g = I_T$.

Proof 2
Take the result Condition for Composite Mapping on Right:

Let $A, B, C$ be sets.

Let $f: B \to A$ and $g: C \to A$ be mappings.

Then:
 * $\operatorname{Im} \left({g}\right) \subseteq \operatorname{Im} \left({f}\right)$

iff:
 * $\exists h: C \to B$ such that $h$ is a mapping and $f \circ h = g$

Let $C = A = T$, let $B = S$ and let $g = I_T$.

Then the above translates into:


 * $\operatorname{Im} \left({I_T}\right) \subseteq \operatorname{Im} \left({f}\right)$

iff:
 * $\exists g: T \to S$ such that $g$ is a mapping and $f \circ g = I_T$

But we know that $\operatorname{Im} \left({f}\right) \subseteq T = \operatorname{Im} \left({I_T}\right)$.

So by definition of set equality, the result follows.

Proof of Non-Uniqueness
If $f$ is not an injection then:
 * $\exists y \in T: \exists x_1, x_2 \in S: f \left({x_1}\right) = y = f \left({x_2}\right)$

Hence we have more than one choice in $f^{-1}\left({\left\{{y}\right\}}\right)$ for how to map $g \left({y}\right)$.

This does not happen iff $f$ is an injection.

Hence the result.

Also see

 * Injection iff Left Inverse