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Dec 27, 2017 at 8:59 vote accept joey112
Dec 26, 2017 at 21:03 comment added Dan Petersen Also, I shouldn't have said "circle" in the comment three days ago, but "2-sphere".
Dec 26, 2017 at 21:01 comment added Dan Petersen The fibration exists, and it is a fiber bundle in the sense of orbifolds. But the fundamental group of the base acts nontrivially on the cohomology of the fiber, so you need to work with twisted coefficients, which kills your suggested argument (the vanishing results for $n=0,1$ are only for constant coefficients).
Dec 26, 2017 at 20:56 answer added Dan Petersen timeline score: 3
Dec 23, 2017 at 11:54 comment added Dan Petersen The space $H_g$ has the rational cohomology of a point. And $H_{g,1}$ has the rational cohomology of a circle. I can supply an argument later.
Dec 23, 2017 at 10:15 review First posts
Dec 23, 2017 at 10:16
Dec 23, 2017 at 10:14 history asked joey112 CC BY-SA 3.0