Bijection iff exists Mapping which is Left and Right Inverse

Theorem
Let $S$ and $T$ be sets.

Let $f: S \to T$ be a mapping.

Then $f$ is a bijection iff:
 * there exists a mapping $g: T \to S$ such that:
 * $g \circ f = I_S$
 * $f \circ g = I_T$
 * where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.

Necessary Condition
Let $f$ be a bijection.

Then for each $y \in T$ there exists one and only one $x \in S$ such that $f \left({x}\right) = y$.

That is, that there exists a mapping $g: T \to S$ with the property that:
 * $\forall y \in T: \exists x \in S: g \left({y}\right) = x$

Let $y \in T$.

Let $x = g \left({y}\right)$.

Then:

and:

demonstrating that $g$ has the property that:
 * $g \circ f = I_S$
 * $f \circ g = I_T$

Sufficient Condition
Suppose there exists a $g$ which satisfies the conditions on $f$.

By Condition for Composite Mapping to be Identity:
 * $(1): \quad$ from $g \circ f = I_S$ it follows that $f$ is an injection.
 * $(2): \quad$ from $f \circ g = I_T$ it follows that $f$ is a surjection.

Thus, by definition, $f$ is a bijection.