Injection iff Left Inverse/Proof 1

Theorem
A mapping $f: S \to T, S \ne \varnothing$ is an injection iff:
 * $\exists g: T \to S: g \circ f = I_S$

where $g$ is a mapping.

That is, iff $f$ has a left inverse.

Proof

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

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

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

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


 * Now, assume $f$ is an injection.

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

As $S \ne \varnothing$, we choose $x_0 \in S$. Then we define:


 * $g \left({y}\right) =

\begin{cases} x_0: & y \in T - \operatorname{Im} \left({f}\right) \\ f^{-1} \left({y}\right): & y \in \operatorname{Im} \left({f}\right) \end{cases} $

Because $f$ is an injection, we know we can do this, as $f^{-1}: \operatorname{Im} \left({f}\right) \to S$ is a mapping, from Injection iff Inverse of Image Mapping.


 * InjectionIffLeftInverse.png

So, for all $x \in S$, $g \circ f \left({x}\right) = g \left({f \left({x}\right)}\right)$ is the unique element of $S$ which $f$ maps to $f \left({x}\right)$.

This unique element is $x$.

Thus $g \circ f = I_S$.

Comment
Notice that it does not matter what the elements of $T \setminus \operatorname{Im} \left({f}\right)$ are - using the construction given, the equation $g \circ f = I_S$ holds whatever value (or values) we choose for $g \left({T \setminus \operatorname{Im} \left({f}\right)}\right)$. The left-over elements of $T$ we can map how we wish and they will not affect the final destination of any $x \in S$ under the mapping $g \circ f$.