# Inverse of Composite Bijection/Proof 2

## Theorem

Let $f$ and $g$ be bijections.

Then:

$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$

and $f^{-1} \circ g^{-1}$ is itself a bijection.

## Proof

Let $f: X \to Y$ and $g: Y \to Z$ be bijections.

Then:

 $\ds \paren {g \circ f} \circ \paren {f^{-1} \circ g^{-1} }$ $=$ $\ds g \circ \paren {\paren {f \circ f^{-1} } \circ g^{-1} }$ Composition of Mappings is Associative $\ds$ $=$ $\ds g \circ \paren {I_Y \circ g^{-1} }$ Composite of Bijection with Inverse is Identity Mapping $\ds$ $=$ $\ds g \circ g^{-1}$ Identity Mapping is Left Identity $\ds$ $=$ $\ds I_Z$ Composite of Bijection with Inverse is Identity Mapping

 $\ds \paren {f^{-1} \circ g^{-1} } \circ \paren {g \circ f}$ $=$ $\ds \paren {f^{-1} \circ \paren {g^{-1} \circ g} } \circ f$ Composition of Mappings is Associative $\ds$ $=$ $\ds \paren {f^{-1} \circ I_Y} \circ f$ Composite of Bijection with Inverse is Identity Mapping $\ds$ $=$ $\ds f^{-1} \circ f$ Identity Mapping is Right Identity $\ds$ $=$ $\ds I_X$ Composite of Bijection with Inverse is Identity Mapping

Hence the result.