3
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ Isn't it true that the forgetful functor $CAlg(C)\to C$ is symmetric monoidal. In that case the functor $CoCAlg(C)\to C$ is symmetric monoidal as well which is what you want. $\endgroup$ – Geoffroy Horel Dec 30 '14 at 11:54
  • 3
    $\begingroup$ Hi Jon, I think Example 3.2.4.4 from HA tells you what you need to know. By the way, there's no Remark 3.4.2.2 in the version that's online now, are you looking at another one? $\endgroup$ – Gijs Heuts Dec 30 '14 at 11:54
  • 2
    $\begingroup$ Whoops, I actually meant 3.2.4.4. But, alright, I'll look more closely at that one. $\endgroup$ – Jonathan Beardsley Dec 30 '14 at 15:04
  • 1
    $\begingroup$ Oh, and hi @GijsHeuts! Hope you're doing well! :-) $\endgroup$ – Jonathan Beardsley Dec 30 '14 at 15:06
  • 1
    $\begingroup$ Jeez, sorry, you guys are right. It's that evaluation functor that Lurie describes in that very Remark. Somehow I was just not reading that part thoroughly. $\endgroup$ – Jonathan Beardsley Dec 30 '14 at 20:49
2
$\begingroup$

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 $C$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.