Composition of Commuting Idempotent Mappings is Idempotent
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Theorem
Let $S$ be a set.
Let $f, g: S \to S$ be idempotent mappings from $S$ to $S$.
Let:
- $f \circ g = g \circ f$
where $\circ$ denotes composition.
Then $f \circ g$ is idempotent.
Proof 1
\(\ds \paren {f \circ g} \circ \paren {f \circ g}\) | \(=\) | \(\ds f \circ \paren {g \circ f} \circ g\) | Composition of Mappings is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds f \circ \paren {f \circ g} \circ g\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {f \circ f} \circ \paren {g \circ g}\) | Composition of Mappings is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds f \circ g\) | $f$ and $g$ are idempotent by hypothesis |
$\blacksquare$
Proof 2
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.
$\blacksquare$