14
$\begingroup$

One has a nice "folk" model structure on strict $\infty$-categories due to Yves Lafont, Francois Metayer and Krzysztof Worytkiewicz whose notion of weak equivalences seem to be the notion of weak equivalences for weak $\infty$-category (I.e. a weaker notion than the existence of a strict inverse).

This produces a weak $(\infty,1)$-category of strict $\infty$-categories. My question is: is it expected to be a full subcategory of the category of weak $\infty$-category ?

(Note: I know that strict infinity categories with strict functors between them do not form a full subcategory of weak infinity categories, what I'm asking here is different, essentially because in the model structure mentioned above not all objects are cofibrant)

I'm actually not sure we have satisfying model for general weak $\infty$-category, and I might prefer to avoid the sort of problems mentioned in this answer, so I'll be happy with an answer dealing with $(\infty,n)$-categories defined for exemple as $n$-fold segal spaces, Rezk $\Theta_n$ spaces, Ara $n$-quasicategories or any other reasonable model. Also an answer focusing on $\infty$-groupoid or $(\infty,1)$-category would already be interesting.

Also, as a side question, assuming this is indeed fully faithful, is there any known result about which are the $\infty$-categories (or maybe $\infty$-groupoids) that are representable by strict $\infty$-categories ?

Edit: Let me clarify a few things which from what I read in the comments where unclear.

From the model structure of Lafont, Metayer, Worytkiewicz one obtains a notion of weak $\infty$-functor between strict $\infty$-category: as every objects in this model structure is fibrant a weak functor (or a weak anafunctor) from $X$ to $Y$ is a morphism $\widetilde{X} \rightarrow Y$ from some cofibrant replacement $\widetilde{X}$ of $X$, and notion of natural isomorphism of weak functor as morphism $\widetilde{X} \rightarrow PY$ where $PY$ is the path object for $Y$ in this model structure.

One can chose a functorial cofibrant replacement to have something more canonical, or even a comonadic one in order to obtain associative composition, but the choice of the cofibrant replacement does not have any effects on the question I'm asking, and it is possible to formulate it without choosing ones.

My question can be formulated as: does it defines the correct set of equivalence class of weak functor between strict $\infty$-categories if one see these as weak $\infty$-categories (and more generally, the correct space of morphism if one push things a little further).

Also note that I'm only interested in the 'canonical' way of sending strict $\infty$-categories to weak $\infty$-category, by just forgeting their strictness.

I know there is ways to send strict $\infty$-categories to weak $\infty$-groupoids or weak $(\infty,1)$-categories by formally (weakly) inverting all arrows or all $k$-arrow for $k>0$ , but then the image by this construction functor is no longer a strict $\infty$-category, and this construction has absolutely no chance to be fully faithful (it will like asking if the geometric realization functor from categories to the homotopy category of spaces is fully faithful).

The construction I'm refering to has good chances to be fully faithful (which is what I'm asking) but is clearly not essentially surjective even on $\infty$-groupoids: the groupoids in its image have for example trivial whitehead products $\pi_2 \times \pi_2 \rightarrow \pi_3$. The follow up question I asked is about knowing if we do have a good characterization of the image of this functor (for example, by the vanishing for all whitehead product in degree higher than $(2,2)$ or something like that). But please don't try to explain that there is construction which allow to represent all $\infty$-groupoid by strict $\infty$-category.

$\endgroup$
  • $\begingroup$ Regarding your last question, I believe that at least all $(\infty,1)$-categories are representable by strict $\infty$-categories. This follows from the Street-Roberts conjecture which was proved by Verity in this paper. Using this conjecture, we can transfer the Joyal model structure on simplicial sets to the category of strict $\infty$-categories and prove that it is a left Bousfield localization of the folk model structure. It follows that the $(\infty,1)$-category of $(\infty,1)$-categories is a full subcategory of strict $\infty$-categories. $\endgroup$ – Valery Isaev Apr 21 '17 at 23:39
  • $\begingroup$ I may have misunderstood what you are asking, but if you consider strict maps between your weak structures then it is true for $\infty$-groupoids (see arxiv.org/abs/1212.3085 ). I suppose you can replicate the same argument for weak $\infty$-categories modeled as in arxiv.org/abs/1009.2331 (though something might go wrong, I haven't thought too hard about it). $\endgroup$ – Edoardo Lanari Apr 22 '17 at 4:10
  • $\begingroup$ @user62782 I think that the presence of a left Bousfield localization in that argument means that it doesn't answer the intended question. If your argument is correct, it shows that weak $(\infty,1)$-categories are equivalent to a full subcategory of strict $(\infty,\infty)$-categories, but that equivalence is not the natural inclusion of strict $(\infty,\infty)$-categories into weak $(\infty,\infty)$-categories. $\endgroup$ – Mike Shulman Apr 22 '17 at 4:48
  • 1
    $\begingroup$ @AndreaGagna I just found my old drafts and it seems there is an error in the argument, so everything I wrote is wrong (probably). Sorry, everyone. $\endgroup$ – Valery Isaev Apr 22 '17 at 13:31
  • 3
    $\begingroup$ I'm not able to follow all the ins and outs of the discussion here. I'll just say that I believe the answer is NO, because the answer for the analogous question about $\infty$-groupoids seems to be no. See my answer at mathoverflow.net/a/225405/437 $\endgroup$ – Charles Rezk Apr 22 '17 at 15:46
2
$\begingroup$

Edit: Previous edit was incorrect.

The naïve answer is no. It follows from Dimitri Ara's paper that the cellular nerve does not preserve fibrations or weak equivalences. Dimitri showed that the cellular nerve of any strict $\omega$-category with a strictly invertible $n$-cell for $n>1$ is not fibrant. To see that weak equivalences are not preserved, notice that the cellular nerve of the polygraphic resolution of $G_2$ where $G_2$ is the strictly contractible $1$-groupoid with two objects has no strictly invertible higher cells and therefore its cellular nerve is fibrant for the Ara-Rezk model structure. However, it can be seen readily that it is not contractible.

The complicated answer is probably yes, in a homotopical sense, but I only have a partial answer for you.

The idea is as follows: Let $C:\operatorname{Cat}_\omega \to \operatorname{Cat}_\omega$ denote the polygraph resolution comonad, and let $\iota:\Theta\hookrightarrow \operatorname{Cat}_\omega$ denote the inclusion functor. Then it is a theorem of Métayer that $C$ is a cofibrant replacement functor for the folk model structure, and it is an observation of Garner that we can compute the pseudofunctors $X\to Y$ by taking the object of morphisms $C(X)\to Y$.

Then the idea, and there is much to check, is as follows:

Let $N_{\mathrm{hc}}: \operatorname{Cat}_\omega \to \widehat{\Theta}$ be the functor defined by the formula $$N_{\mathrm{hc}}(X)_t = \operatorname{Cat}_\omega(C \iota [t], X)$$ for a tree $[t]\in \Theta$.

What isn't too hard to check is the following: The left adjoint to this functor sends boundary inclusions to cofibrations and spine inclusions to trivial cofibrations.

The left adjoint $\mathfrak{C}_{\mathrm{hc}}$ to this functor can be computed as follows:

Let $P:\operatorname{Cat}_\omega\to \operatorname{Poly}$ be the forgetful functor, and let $L: \operatorname{Poly} \to \operatorname{Cat}_\omega$ be the free strict $\omega$-category functor on a polygraph. Then $C=LP$, so to compute $$\mathfrak{C}_{\mathrm{hc}}(A)=\operatorname*{colim}_{[t]\to A} \mathfrak{C}_{hc}([t])=\operatorname*{colim}_{[t]\to A} C\iota[t]=\operatorname*{colim}_{[t]\to A} LP\iota[t]=L(\operatorname*{colim}_{[t]\to A}P\iota [t]),$$ so the the colimit in question can be computed in the category of polygraphs.

In particular, if we take $\partial\Theta[t]\hookrightarrow \Theta[t]$ to be the boundary inclusion, we can see that computing its image under $\mathfrak{C}_{\mathrm{hc}}$ is obtained by applying the functor $L$ to an injective map of polygraphs, which according to Métayer in private correspondence can be shown to be a cofibration.

Then this implies that all injective maps of $\widehat{\Theta}$ are mapped under $\mathfrak{C}_{\mathrm{hc}}$ to cofibrations.

We can also then easily check that the spine inclusion $\operatorname{Sp}[t] \hookrightarrow \Theta[t]$ is a trivial cofibration by noticing that $\mathfrak{C}_{hc}(\operatorname{Sp}[t]) = \iota[t]$, and therefore that $\iota[t]$ is a retract of $\mathfrak{C}_{hc}([t])$ in which the map $\mathfrak{C}_{hc}([t])\to \iota[t]$ is a weak equivalence because $C$ is a cofibrant replacement functor.

Of course, this doesn't prove all we need. We also need to check that $\mathfrak{C}_{hc}$ preserves homotopies, and this becomes very difficult, since polygraphic resolution is extremely wild. I have some ideas to simplify this, but it still looks pretty difficult.

Assuming you can prove that the adjoint pair $\mathfrak{C}_{\mathrm{hc}} \dashv N_\mathrm{hc}$ is a Quillen pair, proving that it is homotopy fully-faithful probably requires an in-depth analysis of Garner's pseudofunctor statement, in particular that all maps $N_{\mathrm{hc}}(X)\to N_{\mathrm{hc}}(Y)$ can be computed as maps $C(X)\to Y$ up to homotopy. A simple manipulation shows that this question can be reduced also to showing that the counit of the adjunction $\mathfrak{C}_{\mathrm{hc}}N_{\mathrm{hc}} \to \operatorname{id}$ is a weak equivalence, which doesn't seem that outlandish. Indeed, it seems to hold for all of the small examples I try to compute by hand.

$\endgroup$
  • $\begingroup$ I'm not certain, but maybe the use of pseudofunctors can overcome the objection of Rezk. $\endgroup$ – Harry Gindi Nov 5 '18 at 0:40
  • $\begingroup$ Don't want to bump it again, but everywhere I used 'polygraph', read 'reflexive polygraph', and 'polygraphic resolution' as 'normalized polygraphic resolution'. Otherwise the spine calculation fails. $\endgroup$ – Harry Gindi Nov 5 '18 at 2:17
  • 1
    $\begingroup$ Interesting, this does not answer the question, but it definitely makes it more tractable. Note however that the expected answer (see the comment of Charles Rezk) is that it is not fully faithful. But it is somehow expected that if you only check low dimensional example and stay away of all kind of "Eckamann-Hilton argument" then you should not see the problem. The first example where this counit is not an equivalence should be at least in dimension $3$, with some non-trivial endomorphisms of an identity in dimension 2. Do you think computations are still tractable there ? $\endgroup$ – Simon Henry Nov 5 '18 at 9:09
  • 1
    $\begingroup$ @SimonHenry I think there is (at least in my thinking, although I have no proof at present) a relatively simple way to demonstrate that the counit is an equivalence whenever $X$ is a finite loop-free pasting in the sense of Steiner (or perhaps more generally when X is the realization of a regular polygraph in the sense of your and Amar's papers, although I am not certain for regular polygraphs). This seems to indicate that the cases where the failure might arise will probably be very complicated, essentially wherever there are nontrivial homotopies. $\endgroup$ – Harry Gindi Nov 5 '18 at 9:34
  • $\begingroup$ Well, conjecturally, it should be true for all "positive polygraphs" i.e. those polygraphs where source and target of generators are non-identity cells, but that is probably harder ! I would be interested to see a proof of this for Steiner categories already. (You might already know that, but note that Amar's notion of regularity and mine are not exactly equivalent, Amar's notion is a little stronger, though his mains results can probably be extended to my notion as well, but I haven't been able to do so) $\endgroup$ – Simon Henry Nov 5 '18 at 9:52

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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