Injection iff Left Inverse/Proof 2

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
Take the result Condition for Composite Mapping on Left:

Let $A, B, C$ be sets.

Let $f: A \to B$ and $g: A \to C$ be mappings.

Then:
 * $\forall x, y \in A: f \left({x}\right) = f \left({y}\right) \implies g \left({x}\right) = g \left({y}\right)$

iff:
 * $\exists h: B \to C$ such that $h$ is a mapping and $h \circ f = g$.

Let $C = A = S$, let $B = T$ and let $g = I_S$.

Then the above translates into:


 * $\exists h: T \to S$ such that $h$ is a mapping and $h \circ f = g$

iff:
 * $\forall x, y \in S: f \left({x}\right) = f \left({y}\right) \implies I_S \left({x}\right) = I_S \left({y}\right)$

But as $I_S \left({x}\right) = x$ and $I_S \left({y}\right) = y$ by definition of identity mapping, it follows that:


 * $\exists h: T \to S$ such that $h$ is a mapping and $h \circ f = g$

iff:
 * $\forall x, y \in S: f \left({x}\right) = f \left({y}\right) \implies x = y$

which is our result.