Inverse of Composite Bijection

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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.


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