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Apr 13, 2017 at 12:32 comment added Artur Jackson @TimCampion, I agree that this is not so enlightening. I think one should find better examples to get a better feel for the issue.
Apr 5, 2017 at 6:36 comment added Tim Campion This is interesting, but I'm not sure how enlightening it is. As Andreas points out, in order to get this result, you need to change the basic localizer on $\mathsf{Cat}$ you're using. In fact, you need to change it to the localizer $\mathcal{W}_n$ which presents $n$-types on $\mathsf{Cat}$. With this localizer, I presume that the induced Cisinski model structure for the test category $\Delta$ also presents $n$-types, so it's not surprising that the model structure for the local test category $\Delta_{\leq n+1}$ does too.
Apr 5, 2017 at 6:29 history edited Artur Jackson CC BY-SA 3.0
Added reference.
Apr 5, 2017 at 6:15 comment added Artur Jackson Andrea, am I off by one here?
Apr 4, 2017 at 17:06 comment added Andrea Gagna I should add that the class of functors you pullback to presheaves categories in this case is not the class of Thomason weak equivalences described in the OP, but instead a further (left Boudfield) localisation.
Apr 4, 2017 at 16:56 comment added Andrea Gagna It is instead $\Delta_{\leqslant n+1}$. Indeed, both the pairs $(i^*, i_*)$ and $(i_!, i^*)$ are Quillen equivalences for the model structures of $n$-equivalences, where here $i\colon \Delta_{\leqslant n+1} \to \Delta$ is the canonical embedding. This is proven, for instance, in Corollary 9.2.7 of Cisinski's book Les préfaisceaux comme modèles des types d'homotopie .
Apr 1, 2017 at 15:31 comment added Marc Hoyois I don't think I believe this. In general there is a conservative functor (of ∞-categories) $\mathcal C[\mathcal W^{-1}] \to Spaces$ whose image consists of all spaces that are homotopy colimits of $C^{op}$-diagrams of sets. When $C=\Delta_{\leq n}$, these are the spaces of homotopy dimension $\leq n$, and it seems unlikely that there is a conservative functor from $n$-types to spaces of homotopy dimension $\leq n$. What is your reasoning?
Apr 1, 2017 at 5:07 comment added Artur Jackson I'm curious if this fits with Anton's comments!
Apr 1, 2017 at 5:06 history answered Artur Jackson CC BY-SA 3.0