# 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:

- $f^{-1} \circ f = I_S$

and:

- $f \circ f^{-1} = I_T$

where $\circ$ denotes composition of mappings.

## 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

- $f \circ f^{-1} = I_T$

$\blacksquare$