Inverse of Composite Bijection/Proof 2

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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

$\Box$


\(\ds \paren {f^{-1} \circ g^{-1} } \circ \paren {g \circ f}\) \(=\) \(\ds \paren {f^{-1} \circ \paren {g^{-1} \circ g} } \circ f\) 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.

$\blacksquare$


Sources

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