# Surjection if Composite is Surjection

## Theorem

Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings such that $g \circ f$ is a surjection.

Then $g$ is a surjection.

## Proof

Let $g \circ f$ be surjective.

Fix $z \in S_3$.

Now find an $x \in S_1: \map {g \circ f} x = z$.

The surjectivity of $g \circ f$ guarantees this can be done.

Now find an $y \in S_2: f \paren x = y$.

$f$ is a mapping and therefore a left-total relation; which guarantees this too can be done.

It follows that:

 $\ds \map g y$ $=$ $\ds \map g {\map f x}$ $\ds$ $=$ $\ds \map {g \circ f} x$ Definition of Composition of Mappings $\ds$ $=$ $\ds z$ Choice of $x$

$\blacksquare$