Bijection has Left and Right Inverse/Proof 3

Theorem
Let $f: S \to T$ be a bijection.

Let:
 * $I_S$ be the identity mapping on $S$
 * $I_T$ be the identity mapping on $T$.

Let $f^{-1}$ be the inverse of $f$.

Then: where $\circ$ denotes composition of mappings.
 * $f^{-1} \circ f = I_S$ and
 * $f \circ f^{-1} = I_T$

Proof
Let $f$ be a bijection.

By definition, $f$ is a mapping, and hence also by definition a relation.

Hence the result Bijective Relation has Left and Right Inverse applies directly and so:
 * $f^{-1} \circ f = I_S$ and
 * $f \circ f^{-1} = I_T$