# Surjection iff Right Inverse/Proof 2

## Theorem

A mapping $f: S \to T, S \ne \O$ is a surjection if and only if:

- $\exists g: T \to S: f \circ g = I_T$

where:

- $g$ is a mapping
- $I_T$ is the identity mapping on $T$.

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

## Proof

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:

- $\Img g \subseteq \Img f$

- $\exists h: C \to B$ 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:

- $\Img {I_T} \subseteq \Img f$

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

But we know that:

- $\Img f \subseteq T = \Img {I_T}$

So by definition of set equality, the result follows.

$\blacksquare$

#### Axiom of Choice

This proof depends on the Axiom of Choice, by way of Condition for Composite Mapping on Right.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.6$