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Dec 26, 2019 at 13:04 vote accept F.Abellan
Dec 26, 2019 at 13:04 vote accept F.Abellan
Dec 26, 2019 at 13:04
Dec 26, 2019 at 13:03 vote accept F.Abellan
Dec 26, 2019 at 13:04
Dec 23, 2019 at 19:18 answer added Tashi Walde timeline score: 10
Dec 12, 2019 at 21:21 answer added Jeremy Rickard timeline score: 10
Dec 11, 2019 at 11:25 comment added F.Abellan My apologies for the late reply. By 1-groupoidification I mean the free groupoid, i.e. the truncated version of the Kan fibrant replacement. Using the description of the Hom-sets in terms of zig-zags of morphisms one can simplify each zigzag with source and target X to obtain the identity if I am not mistaken.
Dec 11, 2019 at 2:37 comment added R. van Dobben de Bruyn To make @MaxNew's objection more precise: on the core you know that $\operatorname{Aut}(X)$ is still trivial by (2), but you don't know if the category is connected. On the free groupoid you know connectedness by (1), but $\operatorname{Aut}(X)$ could be bigger now.
Dec 10, 2019 at 15:55 comment added Max New By groupoidification do you mean the "core" where you take the isomorphisms that already exist or the "free groupoid" which freely adds inverses to all morphisms (the right and left adjoints respectively to the inclusion of groupoids into categories)?
Dec 10, 2019 at 15:45 history asked F.Abellan CC BY-SA 4.0