The same as Symmetric Monoidal Categories

Question

We shall define $\Sigma_*$ as a symmetric operad. If $\sigma\in \Sigma_k$ and $\phi_i\in \Sigma_j$ for $1\leq i\leq k$, then $\Gamma(\sigma,\langle\phi\rangle_1^k)$ is the element of $\Sigma_j$ defined by $$j_1\coprod\cdots \coprod j_k\xrightarrow{\coprod \phi_i}j_1\coprod\cdots \coprod j_k\xrightarrow{\sigma\langle j_1,\cdots, j_k\rangle}j_{\sigma^{-1}(1)}\coprod \cdots\coprod j_{\sigma^{-1}(k)}.$$ Define $E:\textbf{Set}\to \textbf{Cat}$ to be $E(S):=S$ where $S$ is a set and $E(S)$ is a category where $\forall a,b\in S$, there exists a unique isomorphism $a\xrightarrow{f} b$. We have that $E\Sigma_*$ is an operad enriched in $Cat$. An algebra is a map of operads $\mathcal{O}\to End_X$. Note that $End_\mathcal{C}(n)=\underline{Cat}(\mathcal{C}^n,\mathcal{C})$. How do I prove that the algebras of $E\Sigma_*$ are isomorphic to the category of symmetric monoidal categories?

Answer 1

You don't say what sets the $\Sigma_*$ are. It seems that by $\Sigma_n$ you mean the symmetric group, and therefore you mean the natural symmetric operad structure on the family of symmetric groups. However, your notation was confusing to me, which is why I didn't write an answer until now. It might be better in the future to use more words and explain what you mean.

However, if that is what you mean, then the key observation here is that the algebras of $\Sigma_*$ in $\mathbf{Set}$ are (not necessarily commutative) monoids. (Basically by the same argument I gave in my answer last week to your question here). Now $\mu_n$ is the image of $\newcommand\id{\mathrm{id}}\id_n$.

So really we just need to understand what $E$ is doing to $\Sigma_*$. Well, if $E$ didn't toss in all the isomorphisms, if all $E$ did was produce $n$ objects, we'd be taking the free category, so an algebra would be a family of morphisms $\operatorname{Free}(\Sigma_n)\to [\newcommand\C{\mathcal{C}}\C^n,\C]$, or equivalently a family of functions $\Sigma_n \to [\C^n,\C]$, thought of as a choice of functors $F_\sigma : \C^n\to \C$ for all $\sigma\in\Sigma_n$. If $\mu_n : \C^n\to \C$ is the functor corresponding to the identity, then in particular, $F_\sigma = \sigma \cdot \mu_n$, so all of the information is already carried by the $\mu_n$, and we see that algebras for $\operatorname{Free}(\Sigma_n)$ are equivalent to strict monoidal categories.

So what about the isomorphisms? Well, now we still need to choose functors $F_{\sigma}:\C^n\to \C$ for all $\sigma\in\Sigma_n$, we need to choose compatible natural isomorphisms so that for a fixed $n$, all of the $F_\sigma$ are uniquely isomorphic via these chosen natural isomorphisms, and all the composites are compatible.

In particular if we call $F_{\mathrm{id}_2}$ by $\otimes : \C^2\to \C$, then there is a natural isomorphism $$\tau_{X,Y} : \mu_2(X,Y)\to \mu_2(Y,X) = F_{\begin{pmatrix} 2 & 1 \end{pmatrix}}(X,Y).$$ Since transpositions generate the symmetric group, $\tau$ will generate all the other natural isomorphisms.

Finally, note that $\mu_3=\mu_2(\mu_2,\mu_1)=\mu_2(\mu_1,\mu_2)$, so associativity is still strict. Similarly for the unit.

Thus algebras for $E(\Sigma_*)$ are equivalent to symmetric monoidal categories for which the associators and unitors are identities. I.e., a permutative category.