Injection iff Left Inverse/Proof 3

Proof
Let $f: S \to T$ be an injection.

Then $f$ is a one-to-many relation.

By Inverse of Many-to-One Relation is One-to-Many, $f^{-1}: T \to S$ is many-to-one.

By Many-to-One Relation Extends to Mapping, there is a Mapping $g: T \to S$ such that $f^{-1} \subseteq g$.

Let $\tuple {x, y} \in g \circ f$.

Then:
 * $\exists z \in T: \tuple {x, z} \in f, \tuple {z, y} \in g$

Since $\tuple {x, z} \in f$:
 * $\tuple {z, x} \in f^{-1} \subseteq g$

Since $\tuple {z, y} \in g$, $\tuple {z, x} \in g$ and $g$ is a mapping:
 * $x = y$

so:
 * $\tuple {x, y} \in I_S$

So we see that:
 * $g \circ f \subseteq I_S$

Let $x \in S$.

Then:
 * $\tuple {x, \map f x} \in f$

and:
 * $\tuple {\map f x, x} \in f^{-1} \subseteq g$

So:
 * $\tuple {x, x} \in g \circ f$

Thus:
 * $I_S \subseteq g \circ f$

By definition of set equality:
 * $I_S = g \circ f$