# 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.

## Proof

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.

$\blacksquare$