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Sep 4, 2021 at 22:35 vote accept Tim Campion
Sep 4, 2021 at 15:02 comment added Benjamin Steinberg @ChrisSchommer-Pries thanks for the explanation. Ill try to assimilate. My category knowledge stops pretty much at the surface of 2-categories.
Sep 4, 2021 at 14:40 comment added Chris Schommer-Pries Doing that for all the higher simplices too solves the first problem. The second problem is that this doesn't quite work the way we want for basically the same reason that Zhen Lin's suggestion didn't quite work. We can fix it in a way that is morally similar to what Reid Barton suggested, which is to use a slightly bigger model. So you actually need a double subdivision. This double subdivision shows up in Thomason's original work too. Some of the nuance of the Barwick-Kan work is figuring exactly which arrows to localize in the double subdivision.
Sep 4, 2021 at 14:36 comment added Chris Schommer-Pries (cont) If we mark all the backwards arrows as weak equivalences, then when we localize we get something where the arrows can be composed as in the original category. So this is one step closer to what we want. However there are two problems. The first is that we haven't taken care of the compositions in the original category. To handle this we need to use the 2-simplicies in the nerve and use their barycentric subdivision. This will be another poset with certain arrows marked (if you localize those arrows it is equivalent to the usual "2-simplex category" 0 --> 1 --> 2. (cont....)
Sep 4, 2021 at 14:32 comment added Chris Schommer-Pries @BenjaminSteinberg What these constructions do, more or less, is take a category, pass to the nerve, take the subdivision of the nerve, and then use that to glue back some simple categories together to get a poset (and these simple categories have certain arrows marked as "to-be-inverted"). For example, let's take a look at what the subdivision does to an arrow. It replaces the arrow ---> with a pair of arrows pointing towards each other ---> <--- . If we do this replacement to all the arrows in our category, then we get a new category where there are no non-trivial compositions - a poset.
Sep 4, 2021 at 12:36 comment added Benjamin Steinberg As a noncategory theorist why does barycentric subdivision of simplicial sets realize categories as localizations? Do you just replace a category by its nerve?
Sep 4, 2021 at 6:23 comment added Marc Hoyois The variant of the construction given in the exercise is functorial in the simplicial set $S$, but it does not invert categorical equivalences.
Sep 4, 2021 at 2:49 comment added Benjamin Steinberg So is the way you get a category equivalent to my favorite monoid is you use the inverted maps to make objects isomorphic rather than to create automorphisms and you create endomorphisms by forcing certain objects to get identified?
Sep 3, 2021 at 22:05 history became hot network question
Sep 3, 2021 at 21:30 comment added Tim Campion @MarcHoyois Thanks! Do you happen to have studied Lurie's construction of $(P,W)$ closely enough to say whether it's functorial in $C$?
Sep 3, 2021 at 19:29 comment added Peter LeFanu Lumsdaine @MarcHoyois: That seems worth making an answer — it gives a much more self-contained presentation than Chris’s answer.
Sep 3, 2021 at 17:50 comment added Marc Hoyois The first question is answered constructively in Kerodon.
Sep 3, 2021 at 15:59 comment added Reid Barton You need something a bit bigger to get distinct parallel maps in the homotopy category. For example, there's a poset with four objects and four nonidentity maps whose geometric realization is a circle. Now invert three of those maps.
Sep 3, 2021 at 15:52 answer added Chris Schommer-Pries timeline score: 26
Sep 3, 2021 at 15:28 comment added Tim Campion @ZhenLin Hmmm.... If you localize at $0 \to 1$ and $0 \to 2$, then by 2/3 aren't you also localizing at $1 \to 2$, so that you get the terminal category?
Sep 3, 2021 at 15:15 comment added Zhen Lin @TimCampion Isn't $B \mathbb{N}$ the localisation of $\{ 0 \to 1 \to 2 \}$ w.r.t. $\{ 0 \to 1, 0 \to 2 \}$?
Sep 3, 2021 at 14:42 comment added Tim Campion @BenjaminSteinberg Good question. Maybe even $B \mathbb N$ is problematic...
Sep 3, 2021 at 14:39 comment added Benjamin Steinberg Like take the set of all selfmaps of the natural numbers and add an adjoined identity. I don't see how you get this from inverting morphisms in a poset
Sep 3, 2021 at 14:38 comment added Benjamin Steinberg How would you get a monoid with no nonidentity isomorphisms but lots of noninvertible elements this way?
Sep 3, 2021 at 14:16 history edited Tim Campion CC BY-SA 4.0
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Sep 3, 2021 at 14:02 history asked Tim Campion CC BY-SA 4.0