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Sep 12, 2020 at 17:21 vote accept Dmitri Pavlov
Sep 11, 2020 at 22:11 comment added Dmitri Pavlov @TimCampion: Yes, I forgot to include “fibrant” in the original statement, but you already added it.
Sep 11, 2020 at 20:02 history edited Tim Campion CC BY-SA 4.0
As written, the answer would trivially be that this is impossible.
Sep 11, 2020 at 19:43 comment added Tim Campion Regarding the Kan model structure: since every homotopy type is the classifying space of a category, every simplicial is in particular weakly equivalent to a 2-coskeletal simplicial set. Better yet, every homotopy type is the classifying space of a poset, so every simplicial set is weakly equivalent to a 1-coskeletal simplicial set. But of course, a Kan complex which is $n$-coskeletal is $n$-truncated.
Sep 11, 2020 at 18:51 answer added Tim Campion timeline score: 5
Sep 11, 2020 at 18:18 comment added Tim Campion Emily Riehl has shown that Dugger-Spivak mapping spaces are always 3-coskeletal, and that the homspaces of $\mathfrak C X$ are 2-coskeletal when $X$ is a 1-category.
Sep 11, 2020 at 18:15 comment added Phil Tosteson Maybe you can promote your Kan example into a Joyal example by taking the simplicial category with two objects $a,b$ and $Hom(a,b) = N(M)$ (self maps are just the identity), and then applying the simplicial nerve to get a simplicial set. Haven't checked whether this works.
Sep 11, 2020 at 18:07 comment added Tim Campion Up to $\infty$-categorical equivalence, the $n$-coskeletal quasicategories are precisely the quasicategories with $(n-1)$-truncated mapping spaces. So a good place to start would be to ask to slightly weaker question "What are some 2-coskeletal simplicial sets which are not the nerves of 1-categories?"
Sep 11, 2020 at 17:34 history asked Dmitri Pavlov CC BY-SA 4.0