Inverse of Composite Bijection
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Theorem
Let $f$ and $g$ be bijections such that $\Dom g = \Cdm f$.
Then:
- $\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$
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 $f: X \to Y$ 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
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Exercise $\text{J}$
- 2011: Robert G. Bartle and Donald R. Sherbert: Introduction to Real Analysis (4th ed.) ... (previous) ... (next): $\S 1.1$: Sets and Functions