Bijection has Left and Right Inverse/Proof 2

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
Suppose $f$ is a bijection.

From Bijection iff Inverse is Bijection and Bijection Composite with Inverse, it is shown that the inverse mapping $f^{-1}$ such that: is a bijection.
 * $f^{-1} \circ f = I_S$
 * $f \circ f^{-1} = I_T$