Timeline for Grothendieck's Tohoku Paper and Combinatorial Topology
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
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| Jun 21, 2021 at 1:34 | comment | added | Andy Putman | Just noticed this ancient answer. See Prop 1.1 of these notes of mine for a simple and direct proof (less fancy than Oscar’s): www3.nd.edu/~andyp/notes/FiniteOrderHomology.pdf | |
| Feb 20, 2017 at 3:03 | comment | added | Dan Ramras | @jdc The result in Hatcher only covers the case in which the action of the finite group $G$ is free (which in that case is equivalent to the quotient map being a covering space projection). The result I'm quoting doesn't assume the action is free. | |
| Feb 19, 2017 at 6:47 | comment | added | jdc | Very late addition: Unless I'm missing something in what you intend, this has a short, very elementary proof in singular cohomology, making use of the transfer homomorphism. This is in Hatcher's algebraic topology text as Proposition 3G.1, page 321. | |
| Sep 28, 2010 at 0:27 | comment | added | Dan Ramras | HYYY: you just need to know that the cohomology can be calculated using sheaf cohomology; that is, by taking an injective resolution of the constant sheaf. If $X$ is paracompact Hausdorff, that's the same as Cech cohomology. If $X$ has the homotopy type of a CW complex, it's Cech cohomology agrees with its singular cohomology (note also that CW complexes are paracompact and Hausdorff). Finally, compact manifolds always have the homotopy type of CW complexes. (I think the question of whether non-compact manifolds have the htpy type of CW complexes came up on MO recently... I forget where.) | |
| Sep 27, 2010 at 11:40 | comment | added | HYYY | By the way, is it still true for X being a manifold? | |
| Sep 18, 2010 at 0:50 | comment | added | Dan Ramras | HYYY: Oscar suggests (above) looking in the "Alaska notes" as those Proceedings by May et. al. are known. I haven't done so, and I don't know anywhere else to look. | |
| Sep 17, 2010 at 21:58 | comment | added | HYYY | if that is already in equivariant homotopy theory...? | |
| Sep 17, 2010 at 21:57 | comment | added | HYYY | @Dan, hi,do you know if that result has been written down somewhere except in the Tohoku paper? | |
| Sep 17, 2010 at 15:56 | history | edited | Dan Ramras | CC BY-SA 2.5 |
included precise reference
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| Sep 17, 2010 at 15:54 | comment | added | Dan Ramras | Agusti, I've edited to include this information. | |
| Sep 17, 2010 at 8:27 | comment | added | Agustí Roig | @Dam. It's been a while since you posted this answer, but I read it right now and I'm interested: which result is this in the Tohoku paper? | |
| Jul 4, 2010 at 19:18 | comment | added | Oscar Randal-Williams | I don't think I've seen this anywhere, but presumably should be covered in any textbook on equivariant homotopy theory. I've seen similar things in "Equivariant Homotopy and Cohomology Theory", by May et al. | |
| Jun 30, 2010 at 14:29 | comment | added | Dan Ramras | Interesting. I've never seen that before. Is it written down anywhere? I happen to be using this fact in a paper at the moment, which is how I stumbled across this question. | |
| Jun 30, 2010 at 9:24 | comment | added | Oscar Randal-Williams | One can see this by considering the map $X \times_G EG \to X/G$ from the Borel construction, filtering both sides by skeleta of $X$, and calculate directly that it is true for wedges of spheres (for which it is enough to treat the case where $G$ acts transitively on the spheres: then the actual quotient is a sphere, and the homotopy quotient is a sphere times the classifying space of the stabiliser, a finite group). | |
| Jun 30, 2010 at 6:52 | history | answered | Dan Ramras | CC BY-SA 2.5 |