# Identity Mapping is Idempotent

## Theorem

Let $S$ be a set.

Let $I_S: S \to S$ be the identity mapping on $S$.

Then $I_S$ is idempotent:

$I_S \circ I_S = I_S$

## Proof

$I_S \circ f = f$

for all mappings $f: S \to S$.

$f \circ I_S = f$

for all mappings $f: S \to S$.

Substituting $I_S$ for $f$ in either one and the result follows.

$\blacksquare$