Timeline for Homotopy fiber of a map between classifying spaces
Current License: CC BY-SA 4.0
15 events
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Jun 24 at 22:28 | history | edited | Omar Antolín-Camarena | CC BY-SA 4.0 |
Fixed link to proof
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Feb 28, 2018 at 6:47 | history | edited | Omar Antolín-Camarena | CC BY-SA 3.0 |
updated broken URL
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Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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Apr 22, 2016 at 22:48 | vote | accept | Omar Antolín-Camarena | ||
Apr 7, 2016 at 18:23 | history | edited | Omar Antolín-Camarena | CC BY-SA 3.0 |
added guess for missing hypothesis
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Apr 6, 2016 at 23:56 | answer | added | Mike-Doherty | timeline score: 4 | |
Apr 6, 2016 at 22:01 | history | edited | Omar Antolín-Camarena | CC BY-SA 3.0 |
added 296 characters in body
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Apr 6, 2016 at 21:58 | comment | added | Omar Antolín-Camarena | Well, all definitions should be equivalent so I don't really see how it matters, @ViditNanda, but let's say the homotopy fiber over $y_0$ of a map $g : X \to Y$ is the (homotopy type of the) space $\{ (x,p) \in X \times Y^{[0,1]} : p(0) = y_0, p(1) = g(x)\}$. The point is that it is a notion defined for arbitrary maps between spaces, if in the special case of a map $Bf : BG \to BH$ it admits an equivalent description as a classifying space of a category, then I'd say that requires some argument. | |
Apr 6, 2016 at 21:39 | comment | added | Vidit Nanda | @OmarAntolín-Camarena So just to confirm, what is your preferred definition of the homotopy fiber of $Bf$? | |
Apr 6, 2016 at 21:08 | comment | added | Omar Antolín-Camarena | @ViditNanda: Well, (1) it does follow from that, but you'd need to prove that :); (2) I think your characterization is easy to prove if G and H are discrete, is it also easy in general? | |
Apr 6, 2016 at 20:58 | comment | added | Vidit Nanda | I don't have a reference, but doesn't an answer to the first question follow from the fact that the homotopy fiber of $Bf$ is the classifying space of a category whose objects are elements of $H$ and morphisms $h \to h'$ consist of $g \in G$ so that $f(g)h' = h$? | |
Apr 6, 2016 at 20:53 | comment | added | Omar Antolín-Camarena | Borel sounds pretty likely, @RyanBudney. | |
Apr 6, 2016 at 20:38 | comment | added | Ryan Budney | I'm interested in the answers as well but I imagine they all go back to at least Borel, perhaps earlier. | |
Apr 6, 2016 at 19:15 | history | edited | Omar Antolín-Camarena | CC BY-SA 3.0 |
deleted 13 characters in body
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Apr 6, 2016 at 18:48 | history | asked | Omar Antolín-Camarena | CC BY-SA 3.0 |