Bijection has Left and Right Inverse/Proof 1

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.

Then it is both an injection and a surjection, thus both the described $g_1$ and $g_2$ must exist from Injection iff Left Inverse and Surjection iff Right Inverse.

Now we need to show that $g_1 = g_2$.

Thus: