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. 



Opposite Symmetric Monoidal Structure on an Infinity Category
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](http://arxiv.org/pdf/1409.2165.pdf) 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 $C$. 


