The same as the category of commutative monoids.

Question

Let $*:*(n)=*$. A map of operads $$*\to End_A$$ gives us for all $n\in \mathbb{N}$ a map $$*\to Set(A^n,A).$$ In particular, $$*\to Set(A^2,A)$$ gives us a selected function $\mu:A^2\to A$. How do I prove that algebras over the operad $*$ in sets are precisely commutative monoids?

Answer 1

I'll start off by copying my comment, since comments aren't permanent.

Since you don't specify, my first thought was that you were talking about planar operads, but then we get all monoids, not just the commutative ones, so I assume you want symmetric operads.

Note that a symmetric operad has an action of $S^n$ in degree $n$, which we can think of as permuting the variables of the $n$-ary operations.

So an operad morphism $*\to \operatorname{End}(A)$ is a choice $\mu_n$ for all $n\ge 0$ of a symmetric function $A^n\to A$ subject to the constraint that all compositions are compatible and $\mu_1$ is necessarily the identity map.

In particular, if we start with a commutative monoid $(A,*,1)$, then define $$\mu_n(a_1,\ldots,a_n) = \prod_{i=1}^n a_i,$$ this gives a map $*\to \operatorname{End}(A)$.

Conversely, given the maps $\mu_i$, we have $\mu_2(a,b)=\mu_2(b,a)$ by symmetry. Moreover $$\mu_2(a,\mu_0) = \mu_2(\mu_1(a),\mu_0)= \mu_1(a)=a,$$ so $\mu_0$ is an identity for $\mu_2$. Finally $$\mu_3(a,b,c) = \mu_2(a,\mu_2(b,c)) = \mu_2(\mu_2(a,b),c),$$ so $\mu_2$ is associative. Thus $(A,\mu_2,\mu_0)$ is a commutative monoid.