# Inverse of Composite Bijection

## Theorem

Let $f$ and $g$ be bijections.

Then:

$\left({g \circ f}\right)^{-1} = f^{-1} \circ g^{-1}$

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

## Proof 1

$\left({g \circ f}\right)^{-1} = f^{-1} \circ g^{-1}$ is a specific example of Inverse of Composite Relation.

As $f$ and $g$ are bijections then by Bijection iff Inverse is Bijection, so are both $f^{-1}$ and $g^{-1}$.

By Composite of Bijections is Bijection, it follows that $f^{-1} \circ g^{-1}$ is a bijection.

$\blacksquare$

## Proof 2

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

Then:

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

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

Hence the result.