# Equivalence of Definitions of Involutive Mapping

## Theorem

The following definitions of the concept of Involutive Mapping are equivalent:

### Definition 1

$f: A \to A$ is an involution precisely when:

$\forall x \in A: \map f {\map f x} = x$

That is:

$f \circ f = I_A$

where $I_A$ denotes the identity mapping on $A$.

### Definition 2

$f: A \to A$ is an involution precisely when:

$\forall x, y \in A: \map f x = y \implies \map f y = x$

### Definition 3

Let $f: A \to A$ be a mapping on $A$.

Then $f$ is an involution if and only if $f$ is both a bijection and a symmetric relation.

That is, if and only if $f$ is a bijection such that $f = f^{-1}$.

## Proof

### $(1)$ implies $(2)$

Let $f: A \to A$ be an involutive mapping by definition 1.

Then by definition:

$(1): \quad \forall x \in A: \map f {\map f x} = x$

Let $\map f x = y$.

Then by substituting $y$ for $\map f x$ into $(1)$:

$\forall x \in A: \map f y = x$

Thus $f: A \to A$ is an involutive mapping by definition 2.

$\Box$

### $(2)$ implies $(1)$

Let $f: A \to A$ be an involutive mapping by definition 2.

Then by definition:

$\forall x, y \in A: \map f x = y \implies \map f y = x$

and so substituting $\map f x$ for $y$ we have:

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

Thus $f: A \to A$ is an involutive mapping by definition 1.

$\Box$

### $(1)$ iff $(3)$

This is demonstrated in Mapping is Involution iff Bijective and Symmetric.

$\blacksquare$