Opposite Symmetric Monoidal Structure on an Infinity Category

Question

Given an $\infty$-category (in the sense of Lurie) $C$, and a symmetric monoidal structure on $C$ associated to a coCartesian fibration $p:C^\otimes\to N(Fin_\ast)$, Lurie says in Remark 2.4.2.7 of Higher Algebra that by identifying $C$ with a functor $N(Fin_\ast)\to Cat_\infty$ and composing with the involution $R:Cat_\infty\to Cat_\infty$, $C\mapsto C^{op}$, we can induce a symmetric monoidal structure on $C^{op}$ induced from that of $C$ which is unique up to contractible ambiguity (I'm fudging some details here, like the fact that this involution is actually defined on a different but equivalent category).

My question is the following: Let $C$ be a symmetric monoidal category. I want to consider the category of commutative algebras in $C^{op}$ (using the induced symmetric monoidal structure described above), $CAlg(C^{op})$. This category admits a symmetric monoidal structure of its own (see Remark 3.2.4.4 of Lurie's Higher Algebra). Thus, we can again take opposites to obtain a symmetric monoidal category $CoCAlg(C)\equiv(CAlg(C^{op})^{op})$. Let $\star$ denote the finally induced symmetric monoidal structure on $CoCAlg(C)$. Is it true that for $X$ and $Y$ objects of $CoCAlg(C)$, $X\star Y\simeq X\otimes Y$ where $\otimes$ is the original symmetric monoidal structure on $C$? If so, is there an easy way to prove this? It seems like a pretty reasonable thing to expect.

Answer 1

First I think it's important to notice one thing: if there is a cocartesian fibration of $\infty$-operads $C^\otimes\to Fin_\ast$ determining a symmetric monoidal structure on an $\infty$-category $C$, it is not the opposite fibration which determines the symmetric monoidal structure on $C^{op}$. In other words, there really isn't a cocartesian fibration $C^{\otimes,op}\to Fin_\ast$ that does what we want. The way to get this is to use a recent paper of Barwick, Glassman and Nardin that produces the dual cartesian fibration $C^{\otimes,\vee}\to Fin_\ast^{op}$ and then apply $op$, to obtain a cocartesian fibration $(C^{\otimes,\vee})^{op}\to Fin_\ast$ (sorry for all of the decorations). Anyway, it's this fibration whose fiber over $\langle n\rangle$ is $(C^{op})^n$ and which classifies the composite functor $op\circ C^\otimes:Fin_\ast\to Cat_\infty$.

Anyway, following up with Gijs, Sanath and Geoffrey's comments, I'll say the following as a final answer:

Following Remark 3.2.4.4 of Higher Algebra, we have that if $C$ is symmetric monoidal then the forgetful functor $U:CAlg(C)^\otimes\to C$ (given by evaluation on $\langle 1\rangle\in Fin_\ast$) is itself a symmetric monoidal functor, indicating that if $\Box$ is the symmetric monoidal structure on $Alg(C)$ then $U(A\Box B)\simeq A\otimes B$, where $\otimes$ is the symmetric monoidal structure on $C$. So, for us, this means that there is a forgetful functor whcih is also symmetric monoidal $CAlg(C^{op})\to C^{op}$ which which shows that tensor products in $CAlg(C^{op})$ are computed the same way tensor products in $C^{op})$ are.

This of course doesn't quite give us the solution rigorously, since there's no way to get from $C^{op}$ back to $C$, and we would like an honest equivalence. Thus, we have to use the fact that since we have put a symmetric monoidal structure on $CAlg(C^{op})$ given by $CAlg(C^{op})^\otimes\to Fin_\ast$, we can produce a cocartesian fibration (again using the technology described above) $(CAlg(C^{op})^{\otimes,\vee})^{op}\to Fin_\ast$ whose fiber over $\langle n\rangle$ is $(CAlg(C^{op})^{op})^n$. Then we note that there is a forgetful functor down to $C^\otimes$ and since none of the above constructions change any of the important structure fiberwise, we have an equivalence of the forgetful map applied to the tensor of two coalgebras with their tensor in $C$.