Involution is Permutation

Theorem

Let $S$ be a set.

Let $f: S \to S$ be an involution.

Then $f$ is a permutation.

Proof

By definition, a permutation is a bijection from a set to itself.

Thus it is sufficient to show that $f$ is a bijection.

By definition of involution, for each $x \in S$:

$\map f {\map f x} = x$

By Equality of Mappings:

$f \circ f = I_S$

where $I_S$ is the identity mapping on $S$.

Thus $f$ is both a left inverse and a right inverse of itself.

The result follows from Bijection iff Left and Right Inverse.

$\blacksquare$