# Injection iff Left Inverse/Proof 1

## Theorem

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

$\exists g: T \to S: g \circ f = I_S$

where $g$ is a mapping.

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

## Proof

Let:

$\exists g: T \to S: g \circ f = I_S$

From Identity Mapping is Injection, $I_S$ is injective, so $g \circ f$ is injective.

So from Injection if Composite is Injection, $f$ is an injection.

Note that the existence of such a $g$ requires that $S \ne \O$.

$\Box$

Now, assume $f$ is an injection.

We now define a mapping $g: T \to S$ as follows.

As $S \ne \O$, we choose $x_0 \in S$.

By definition of injection:

$f^{-1} {\restriction_{\Img f} } \to S$ is a mapping

so it is possible to define:

$\map g y = \begin{cases} x_0: & y \in T \setminus \Img f \\ \map {f^{-1} } y: & y \in \Img f \end{cases}$

It does not matter what the elements of $T \setminus \Img f$ are.

Using the construction given, the equation $g \circ f = I_S$ holds whatever value (or values) we choose for $g \sqbrk {T \setminus \Img f}$.

The remaining elements of $T$ can be mapped arbitrarily, and will not affect the image of any $x \in S$ under the mapping $g \circ f$.

So, for all $x \in S$:

$\map {g \circ f} x = \map g {\map f x}$

is the unique element of $S$ which $f$ maps to $\map f x$.

This unique element is $x$.

Thus $g \circ f = I_S$.

$\blacksquare$