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Feb 16, 2016 at 7:14 vote accept dorebell
Feb 14, 2016 at 0:45 comment added John Klein @Sebastian Goette: I don't understand your qualm. the construction I have in mind is this: For any space $B$ Milnor gives an explicit construction of a universal principal bundle $U \to B$ with respect to a topological group $G$. Here $U$ is contractible. If $E \to B$ is a any Serre fibration we can form the fiber product $F:= E \times^{B} U$. Then $F$ comes equipped with a (free) $G$-action and the Borel construction $EG\times_G F \to BG$ is a fiber bundle that is fiber homotopy equivalent to the original fibration $E \to B$.
Feb 12, 2016 at 20:29 comment added Sebastian Goette @JohnKlein To the best of my knowledge, a Seifert fibration is an honest fibration in the orbifold sense. But it is not a Serre fibration if we regard the base as a CW complex in the naïve way. If you turn it into a fibre bundle, the fibre can be anything except $S^1$. Indeed, I would be happy to see a good geometric description of the homotopy fibre leading to the correct Leray-Serre spectral sequence.
Feb 12, 2016 at 11:17 comment added John Klein Any Serre fibration over a finite CW base is fiberwise weak equivalent to a fiber bundle. So I don't understand what the issue is....
Feb 12, 2016 at 7:18 answer added Mark Grant timeline score: 5
Feb 12, 2016 at 6:32 comment added Dan Petersen I'm rather ignorant about 3-manifolds, but I thought a Seifert fiber space was an honest-to-God fiber bundle (tho' over an orbifold). And the sheaf-theoretic derivation/interpretation works just as well for orbifolds/stacks.
Feb 12, 2016 at 4:00 comment added dorebell Thanks! In my case, I understand the derived pushforward reasonably well. The thing I don't understand is the edge map itself, especially on the torsion subgroup. I know the form of the sequence, but the Grothendieck formalism makes it hard to see what the ($E_2$) maps are. I want something like the CW-complex answer for fibrations. (If it helps: in my case, all fibers are $S^1$, but they have multiplicity at some points)
Feb 12, 2016 at 3:45 comment added Dylan Wilson Indeed, the Grothendieck spectral sequence tells you that you need to understand the derived pushforward. In good cases (presumably including the one you're talking about) the derived pushforward of the constant sheaf isn't quite a local system but you can stratify the base so that it's a local system on each stratum (I.e. You have a constructible sheaf) and then you can try to compute from there.
Feb 12, 2016 at 3:40 answer added Mike-Doherty timeline score: 1
Feb 11, 2016 at 23:21 comment added Denis Nardin Why don't you want to use the Grothendieck spectral sequence? Intuitively it should generalize much more easily than any topological method. The topological method is usually used only for the special case of fiber bundles and local coefficients (that is locally constant sheaves) and I don't know how much of it can be generalized.
Feb 11, 2016 at 22:53 history asked dorebell CC BY-SA 3.0