Injection if Composite is Injection

Theorem

Let $f$ and $g$ be mappings such that their composite $g \circ f$ is an injection.

Then $f$ is an injection.

Let $g \circ f$ be injective.

We need to show that $f \left({a}\right) = f \left({b}\right) \implies a = b$.

So suppose $f \left({a}\right) = f \left({b}\right)$.

Then:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle g \circ f \left({a}\right)$	$=$	$\displaystyle $	$\displaystyle g \left({f \left({a}\right)}\right)$	$\displaystyle $	$\displaystyle $	Definition of Composite
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle $	$\displaystyle g \left({f \left({b}\right)}\right)$	$\displaystyle $	$\displaystyle $	By Hypothesis
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle $	$\displaystyle g \circ f \left({b}\right)$	$\displaystyle $	$\displaystyle $	Definition of Composite

... thus $a = b$ as $g \circ f$ is an injection.

So we have shown that $f \left({a}\right) = f \left({b}\right) \implies a = b$.

Hence the result from the definition of injection.

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle g \circ f \left({a}\right)\)	\(=\)	\(\displaystyle \)	\(\displaystyle g \left({f \left({a}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	Definition of Composite
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle g \left({f \left({b}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	By Hypothesis
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle g \circ f \left({b}\right)\)	\(\displaystyle \)	\(\displaystyle \)	Definition of Composite