Composition of Commuting Idempotent Mappings is Idempotent/Proof 2
Let $S$ be a set.
Let $f, g: S \to S$ be idempotent mappings from $S$ to $S$.
- $f \circ g = g \circ f$
where $\circ$ denotes composition.
Then $f \circ g$ is idempotent.
By Set of all Self-Maps under Composition forms Semigroup, the set of all self-maps on $S$ forms a semigroup under composition.
The result follows from Product of Commuting Idempotent Elements is Idempotent.