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

$\left({g \circ f}\right)^{-1} = f^{-1} \circ g^{-1}$ 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, it follows that $f^{-1} \circ g^{-1}$ is a bijection.

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 3.6$: Example $53$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 5$: Theorem $5.6$
• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): $\text{I}$: Exercise $\text{J}$
• T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 5$: Theorem $5.10 \ (3)$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 25.2$
• H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability (1996): Appendix $\text{A}.7$: Proposition $\text{A}.7.5 \ (3)$