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
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 3.6$. Products of bijective mappings. Permutations: Example $53$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.6$
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.10 \ (3)$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 25.2$: Some further results and examples on mappings
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.5 \ (3)$