# Composition of Idempotent Mappings

## Theorem

Let $S$ be a set.

Let $f, g: S \to S$ be idempotent mappings.

Suppose that $f \circ g$ and $g \circ f$ have the same images.

That is, suppose that $f \sqbrk {g \sqbrk S} = g \sqbrk {f \sqbrk S}$.

Then $f \circ g$ and $g \circ f$ are idempotent.

## Proof

Let $x \in S$.

By the premise:

$\map f {\map g x} \in g \sqbrk {f \sqbrk S}$

Since $f \sqbrk S \subseteq S$:

$\map f {\map g x} \in g \sqbrk S$

Thus for some $y \in S$:

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

Since $g$ is idempotent:

$\map g {\map g y} = \map g y$

By the choice of $y$:

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

Thus:

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

That is, $f \circ g$ is idempotent.

$\blacksquare$