# Set of all Self-Maps is Monoid

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## Theorem

Let $S$ be a set.

Let $S^S$ be the set of all mappings from $S$ to itself.

Let the operation $\circ$ represent composition of mappings.

Then the algebraic structure $\struct {S^S, \circ}$ is a monoid whose identity element is the identity mapping on $S$.

## Proof

By Set of All Self-Maps is Semigroup, $\struct {S^S, \circ}$ is a semigroup.

By Identity Mapping is Left Identity and Identity Mapping is Right Identity the identity mapping on $S$ is the identity element of $\struct {S^S, \circ}$.

Since $\struct {S^S, \circ}$ is a semigroup with an identity element, it is a monoid by definition.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 4.3$. Units and zeros: Example $73$ - 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 5$ - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 3.1$: Monoids: Exercise $(2)$