Transformation semigroup

Remarks

Composition of maps is associative, so a transformation semigroup is a semigroup. The whole set $T_X$ is the full transformation monoid; for $X$ with $n$ elements it has $n^n$ elements. Its invertible elements form the symmetric group on $X$. Every semigroup $S$ is isomorphic to a transformation semigroup on the set $S^1$, by sending $a$ to the map $x \mapsto ax$.

Example: the constant maps $c_a \colon x \mapsto a$, for $a \in X$. For all $a, b$, $c_a \circ c_b = c_a$, so they form a transformation semigroup.

Non-example: on $X = \set{1, 2}$, let $s$ swap $1$ and $2$, and let $c_1$ be the constant map to $1$. The set $\set{\id, s, c_1}$ is not closed under composition, because $s \circ c_1 = c_2$ is not in it.