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Jan 26, 2017 at 17:42 vote accept KotelKanim
Jan 24, 2017 at 8:51 comment added KotelKanim OK. Now I am convinced. Thank you very much for this observation, it is very nice. Even though I was interested in the strict version, if no one posts an answer regarding that, I will accept this one, since it answers something close and interesting by itself.
Jan 22, 2017 at 16:14 comment added Karol Szumiło By analyzing pushout products of boundary inclusions you can check that the lifting property that I described is equivalent to the (seemingly stronger) condition that $\mathcal{C}^{\Delta[m]} \to \mathcal{C}^{\partial\Delta[m]}$ is an acyclic Kan fibration for $m \ge n + 2$. This is invariant under categorical equivalences.
Jan 22, 2017 at 5:28 comment added KotelKanim I think you are right about the lifting property, I was confused by not taking into account correctly that it is already a quasi-category, but I still lack the precise argument. Every quasi-category is indeed categorically equivalent to the coherent nerve of a fibrant simplicial category, but why is the lifting property condition invariant under categorical equivalence? I am sorry if I am being a bit dense about this...
Jan 21, 2017 at 23:04 comment added Karol Szumiło The strict definition of $(n, 1)$-categories is indeed not invariant under categorical equivalences. If you want to use it to construct a Bousfield localization from it, what model structure would you localize? I don't see any candidate other than the Joyal model structure and if you try that with the Joyal model structure you will end up with the same localization that I described anyway.
Jan 21, 2017 at 23:01 comment added Karol Szumiło Why do you believe that the lifting property I gave is not equivalent to $(n - 1)$-truncated mapping spaces? It is fairly easy to see, e.g., you can assume that your quasicategory is the coherent nerve of a fibrant simplicial category and use the left adjoint $\mathfrak{C}$.
Jan 21, 2017 at 8:24 comment added KotelKanim Anyway, I suppose there is a way to rephrase it using some lifting property, which you can then plug into the Bousfield localization machine. I did hope for a strict version though. As you say, its weak equivalences would be incomparable with categorical equivalences, hence it will not be a localization of the Joyal model structure.
Jan 21, 2017 at 8:17 comment added KotelKanim It is true that an $\infty$-category is equivalent to an $n$-category iff its mapping spaces are $(n-1)$-truncated (homotopically). I don't see why (and I don't think it's true that) it is equivalent to the lifting property you suggest. Being equivalent to a 1-category is not just being 2-co-skeletal, right?
Jan 20, 2017 at 19:52 comment added Karol Szumiło I have just remembered that Rezk considers the Segal space version of this problem in A Cartesian Presentation of Weak n-categories. In fact, he covers general $(n, k)$-categories not just $(n, 1)$-categories. He uses a similar Bousfield localization of the model structure for complete $\Theta_k$-spaces.
Jan 20, 2017 at 18:06 history answered Karol Szumiło CC BY-SA 3.0