Inverse of Inverse of Bijection/Proof 1

Theorem

Let $f: S \to T$ be a bijection.

Then:

$\paren {f^{-1} }^{-1} = f$

where $f^{-1}$ is the inverse of $f$.

Proof

Let $f: S \to T$ be a bijection.

From Composite of Bijection with Inverse is Identity Mapping we have:

$f^{-1} \circ f = I_S$

$f \circ f^{-1} = I_T$

where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.

The result follows from Left and Right Inverses of Mapping are Inverse Mapping.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.5$
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 25.3$: 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 $\text{A}.7.5 \ (2)$