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Mar 20, 2020 at 1:06 vote accept Tim Campion
Feb 29, 2020 at 10:08 comment added Yonatan Harpaz Another proof of this fact using Quillen cohomology of categories and obstruction theory appears in Example 3.3.11 of londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/topo.12074 (there is a free version on my site, but I see that the arxiv version is not up to date though). Roughly speaking, the point is that $Idem$ has low "Quillen cohomological dimension", and so once you know the first few coherences, the rest of the obstructions are zero.
Feb 26, 2020 at 22:49 history edited Tim Campion CC BY-SA 4.0
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Feb 26, 2020 at 22:46 answer added Tim Campion timeline score: 2
Feb 26, 2020 at 20:35 history edited Tim Campion CC BY-SA 4.0
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Feb 26, 2020 at 20:12 history edited Tim Campion CC BY-SA 4.0
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Feb 26, 2020 at 20:05 history edited Tim Campion CC BY-SA 4.0
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Feb 26, 2020 at 18:59 comment added Tim Campion Something is wrong -- a fundamental group / homology computation shows that the weak homotopy type of the $n$-skeleton of $Idem$ is $S^n$ for $n$ odd, and weakly contractible for $n$ even. $Idem$ itself is weakly contractible, and homotopy cofinal functors preserve weak homotopy type, so the inclusion of the $n$-skeleton can only be homotopy cofinal for $n$ even. (This condition is necessary but not sufficient -- e.g. the 0 and 2 -skeleton inclusions are certainly not homotopy cofinal). In particular, the 3-skeleton is not homotopy cofinal in $Idem$, but perhaps the 4-skeleton is.
Feb 26, 2020 at 18:22 history edited Tim Campion CC BY-SA 4.0
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Feb 26, 2020 at 18:04 history edited Tim Campion CC BY-SA 4.0
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Feb 26, 2020 at 17:58 history asked Tim Campion CC BY-SA 4.0