Set of all Self-Maps under Composition forms 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 under Composition forms 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): Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.1$: Monoids: Exercise $(2)$