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.] 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!


1 Answer 1

  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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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