8
$\begingroup$

$\newcommand{\M}{\mathcal{M}}$

Suppose I have a monoidal simplicial model category in which every object is cofibrant $(\M,\otimes,\mathbb{1})$ and I want to look at its underlying monoidal quasicategory, which I'll write as $N(\M)$, the simplicial nerve of $\M$. One way to do this is the following construction (following Variant 4.1.3.17 of Lurie's Higher Algebra):

First take the sub-simplicial monoidal category of fibrant and cofibrant objects of $\M$, denoted $\M^\circ$. This is still a monoidal simplicial category. Now produce its so-called "category of operators," i.e. the free semi-cartesian monoidal category on its underlying simplicial multicategory. To be explicit, do the following:

Let $Multi(\M^\circ)$ be the simplicial multicategory whose objects are the objects of $M^\circ$ and whose "multimapping" objects are $$Mul(\{m_1,\ldots,m_k\},m)=\coprod_{\alpha\in\Sigma_k} Hom_{\M^\circ}(m_{\alpha(1)}\otimes\cdots\otimes m_{\alpha(k)},m).$$ Now produce the category of operators of this multicategory, which I'll write as $(\M^\circ)^\otimes$. This category has as objects pairs $(\langle k\rangle,\{m_1,\ldots,m_k\})$ where $\langle n\rangle$ is the finite pointed set $\{\ast,1,\ldots,k\}$ and $\{m_1,\ldots,m_k\}$ is a finite list of objects of $Multi(\M^\circ)$ (hence a finite list of objects of $\M^\circ$). This category has mapping objects given by $$Hom_{(\M^\circ)^\otimes}((\langle k\rangle,\{m_1,\ldots,m_k\}),(\langle j\rangle\{l_1,\ldots,l_j\}))=\coprod_{f:\langle k\rangle\to\langle j\rangle}\prod_{1\leq r\leq j}Mul(\{m_i\}_{i\in f^{-1}(r)},l_r).$$

This construction probably seems a bit unwieldy, but the point is that this final category $(\M^\circ)^\otimes$ admits a forgetful functor $(\M^\circ)^\otimes\to Ass^\otimes$, the category of operators of the associative operad. In this case, since all of our mapping objects are Kan complexes (this follows from the fact that we took bifibrant objects and that cofibrant objects are closed under tensor product), we get that $N((M^\circ)^\otimes)\to N(Ass^\otimes)$ defines a monoidal structure on the quasicategory $N(M^\circ)$.

Now, my question is about the following: Suppose my simplicial monoidal model category $\M$ has a tensor product that preserves fibrant objects in addition to cofibrant objects. Then take the opposite category of $\M^\circ$, which we'll write $(\M^\circ)^{op}$, which is still a simplicial monoidal category. Moreover, since the tensor product preserves fibrancy, the mapping objects of $((\M^\circ)^{op})^\otimes$ are Kan complexes and so we still have a structure map of quasicategories $N(((\M^\circ)^{op})^\otimes)\to N(Ass^\otimes)$ defining a monoidal structure on $N((\M^\circ)^{op})$.

My question is the following: is the monoidal structure so defined on $N((\M^\circ)^{op})$ equivalent to the monoidal structure on $N(M^\circ)^{op}$ obtained by using the Grothendieck construction to get a functor $N(Ass^\otimes)\to Cat_\infty$, composing with $op$ to get a functor $N(Ass^\otimes)\to Cat_\infty\overset{op}\to Cat_\infty$ and then reversing the Grothendieck construction?

----(EDIT)----

Perhaps a bit of background. The main point of this whole thing here is to take a simplicial monoidal model category and lift strict comonoids and comodules therein to its associated monoidal quasicategory (in the sense of Lurie). These can be relatively easily defined to be strict monoids and modules in the opposite category, but one runs into difficulties using Lurie's set-up to access the monoidal structure on the nerve of the opposite category. When the tensor product preserves fibrancy it seems like the above construction might be a work around. At the very least, it lets you define comodules and comonoids in SOME monoidal quasicategory, but it's not clear that this is equivalent to the canonically defined (up to contractible space of choices) opposite monoidal quasicategory. If anyone has some other idea for lifting comonoids and comodules, I'd be very interested to know about it.

$\endgroup$
  • 1
    $\begingroup$ Dear Jon, I'm pretty sure that the answer is yes and that the easiest way to prove this would be before taking nerves of anything: there should still be a (monoidal 2-)functor $Ass^\otimes \to Cat_\Delta$ expressing the monoidal structure on $\mathcal{M}$, and the composite with the "opposite" construction classifies the monoidal structure on $\mathcal{M}^{op}$. $\endgroup$ – Tyler Lawson Mar 27 '17 at 0:21
  • 1
    $\begingroup$ If you do it that way, then it is legitimate to just compute mapping spaces in $\mathcal{M}$ and its opposite. If you don't, then I suspect that you need to invoke some of the machinery of arxiv.org/pdf/1409.2165.pdf to switch your cocartesian fibration $p$ to $N(Ass^\otimes)$ to a cocartesian fibration $(p^\vee)^{op}$ classifying the composite with the "op" functor. $\endgroup$ – Tyler Lawson Mar 27 '17 at 0:23
  • $\begingroup$ @TylerLawson Agreed! One concern I have however is that there doesn't seem to be a good description of that kind of enriched Grothendieck construction. Perhaps I should spend some time writing it down though. Dai Tamaki's construction doesn't seem to quite do the job. $\endgroup$ – Jonathan Beardsley Mar 27 '17 at 0:42
3
$\begingroup$

The answer to this question is yes, and it's the main result of this paper.

One thing to point out is that, even in the case that the tensor product of $\mathcal{M}$ preserves fibrant objects (so that the homotopy types of the mapping objects in the multicategory associated to $\mathcal{M}^\circ$ are well-behaved), we don't necessarily get a strictly monoidal simplicial category, which is the input described in the above cited paper. However, that's not a problem as we can apply standard machinery to replace any monoidal simplicially enriched category (i.e. a quasimonoid in the 2-category of simplicially enriched categories) by a strict monoid which has the same underlying quasicategory.

$\endgroup$

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.