7
$\begingroup$

There are various models of $\infty$-categories floating around, so there are as many models of the associated homotopy 1- and 2-categories. Because the relations between the former are worked out in great detail, it feels like the relations between the latter should be well-understood as well. But since I was unable to find much about the 2-dimensional case I'd like to ask for references here.

Specifically I am interested in the following questions.

  1. Given a Kan-enriched category $\mathcal{C}$ we can apply the homotopy coherent nerve to get a quasicategory $N^\Delta(\cal{C})$. We can then apply [HTT Prop. 2.3.4.12] to pass to a simplicial set $h_2(N^\Delta(\cal{C}))$, which Lurie calls the underlying 2-category. On the other hand we can take $\cal{C}$ and apply hom-wise the homotopy-category functor and obtain a Grpd-enriched category, say $H_2(\cal{C})$. We can now apply the Duskin-nerve to get another simplicial set $N^D(H_2(\cal{C}))$, which ought to be a 2-category in the sense of HTT. Are both sSets equivalent in an appropriate sense (I guess it means Joyal-equivalent)?

Even more important to me is the following variation:

  1. Note that $H_2(\cal{C})$ is a Grpd-enriched hence a strict (2,1)-category. In the light of Kerodon Rem. 2.3.6 it seems like the Duskin-nerve has a left adjoint $|-|^D:\mathsf{sSet} \rightarrow (2,1)\mathsf{Cat}_{str}$. So are the strict (2,1)-categories $H_2(\cal{C})$ and $|h_2(N^\Delta(\cal{C}))|^D$ equivalent as strict (2,1)-categories or at least as weak (2,1)-categories?

Out of curiosity I'd like to add a third variation on the question:

  1. Using the Cordier-realization $|-|^C:\mathsf{sSet} \rightarrow \mathsf{sSet}\text{-}\mathsf{Cat}$ and the fibrant replacement of simplicial categories $R:\mathsf{sSet}\text{-}\mathsf{Cat} \rightarrow \mathsf{Kan}\text{-}\mathsf{Cat}$ we can turn a quasicategory $C$ into a Kan-enriched category $R(|C|^C)$. So how are $|h_2(C)|^D$ and $H_2(R(|C|^C))$ related?

Thank you very much for your time and considerations!

$\endgroup$

1 Answer 1

8
$\begingroup$
  1. The simplicial sets $h_2(N^\Delta(\mathcal{C}))$ and $N^D(H_2(\mathcal{C}))$ are isomorphic. To prove this, observe that the universal property of $h_2(N^\Delta(\mathcal{C}))$ applied to the image under $N^{\Delta}$ of the quotient simplicial functor $\mathcal{C} \to H_2(\mathcal{C})$ yields a map of simplicial sets $h_2(N^\Delta(\mathcal{C})) \to N^D(H_2(\mathcal{C}))$, which one can easily check is an isomorphism of simplicial sets (it suffices to check that it's a bijection on 0-, 1-, and 2-simplices).

  2. The strict (2,1)-categories $|h_2(N^\Delta(\mathcal{C}))|^{\mathcal{D}}$ and $H_2(\mathcal{C})$ are biequivalent; in fact, there is a strict 2-functor $|h_2(N^\Delta(\mathcal{C}))|^{\mathcal{D}} \to H_2(\mathcal{C})$ which is a bijective-on-objects biequivalence. To prove this, note that the composite of the Duskin nerve functor (for strict (2,1)-categories) with its left adjoint sends a strict (2,1)-category $\mathcal{A}$ to its "normal pseudofunctor classifier" $Q\mathcal{A}$, which is a strict (2,1)-category with the universal property that strict 2-functors $Q\mathcal{A} \to \mathcal{B}$ are in natural bijection with normal pseudofunctors $\mathcal{A} \to \mathcal{B}$, for $\mathcal{B}$ a strict (2,1)-category. Moreover, by this universal property, there is a "counit" strict 2-functor $Q\mathcal{A} \to \mathcal{A}$ which one can show is bijective on objects and an equivalence on hom-categories, and hence a biequivalence.

  3. The strict (2,1)-categories $|h_2(C)|^D$ and $H_2(R(|C|^C))$ are isomorphic (if you make a good choice of $R$, e.g. change-of-base along $Ex^\infty$). Indeed, the two functors $|h_2(-)|^D$ and $H_2(|-|^C)$ are naturally isomorphic, and the functor $H_2$ sends the "unit" map $\mathcal{E} \to R(\mathcal{E})$ to an isomorphism (for a good choice of $R$ as above).

I don't know any references for these answers, but these are all straightforward and standard arguments. If you would like me to elaborate on any of these points, I would be happy to.

$\endgroup$
1
  • $\begingroup$ Thank you very much, this already helped alot! $\endgroup$ Commented Jan 10, 2022 at 14:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .