Surjection if Composite is a Surjection

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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.

Let $g \circ f$ be surjective.

Fix $z \in S_3$.

Now find an $x \in S_1: g \circ f \left({x}\right) = z$.

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

Let $y = f \left({x}\right)$ (so $y \in S_2$).

It follows that:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle g \left({y}\right)$	$=$	$\displaystyle g \left({f \left({x}\right)}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle g \circ f \left({x}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Definition of Composite
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle z$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Choice of $x$

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle g \left({y}\right)\)	\(=\)	\(\displaystyle g \left({f \left({x}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle g \circ f \left({x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Composite
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Choice of $x$