Showing $(M^M, \circ )$ is a monoid.

Question

Show that group $(M^{M},\circ)$ is a monoid with a neutral element $id_{M}$.

$id_{M}$ is defined as "identity mapping" (closest translation that I could have gotten) $id_{M}: M \rightarrow M, x \mapsto x$, where $M$ is a set and $x$ is an element of a set.

I'm not sure how can I start, especially that I don't really know what set $M^M$ is defined as.

Answer 1

$A^B$ is standard notation for the set of functions with domain $B$ and codomain $A$. Therefore your structure is the set of functions with domain $M$ and codomain$ M$ endowed with the operation of composition of functions.

As for where to start, you really only have the two things to check, which you should be able to do without too much trouble.

Answer 2

Well, the operation is composition. The composition of mappings $f,g\in M^M$ is a mapping $f\circ g\in M^M$. So the operation $\circ : M^M\times M^M\rightarrow M^M:(f,g)\mapsto f\circ g$ is well-defined.

This operation is associative, i.e., for all $f,g,h\in M^M$, $(f\circ g)\circ h = f\circ (g\circ h)$.

Finally, $id_M$ is the identity element, i.e., for all $f\in M^M$, $f\circ id_M = f = id_M\circ f$.