Inverse of Inverse of Bijection

Theorem

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


Then:

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

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


Proof 1

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.

$\blacksquare$


Proof 2

A mapping is a relation.

Thus it follows that Inverse of Inverse Relation can be applied directly.

$\blacksquare$


Sources

  • 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): Exercise $1.3: \ 12 \ \text{(b)}$
  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 3.3$. Injective, surjective, bijective; inverse mappings
  • 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Composition of Functions
  • 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 13$
  • 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.11$: Relations: Theorem $11.10$
  • 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $3$
  • 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
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