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Dec 23, 2010 at 18:10 comment added Sergey Melikhov "How to compute the (co)homology of orbit spaces (when the action is not free)?" mathoverflow.net/questions/1080
S Dec 14, 2010 at 12:35 vote accept HKSHLZW
Dec 14, 2010 at 12:35 vote accept HKSHLZW
S Dec 14, 2010 at 12:35
Dec 14, 2010 at 12:35 vote accept HKSHLZW
Dec 14, 2010 at 12:35
Dec 10, 2010 at 1:00 comment added user2464 Let us try to use a sequence of cutting and you get that you want.
Dec 9, 2010 at 15:34 comment added Dan Ramras For a finite group acting on a reasonable space, the rational cohomology of the quotient is the fixed points in the rational cohomology of the total space (this works with any field coefficients). See my answer here: mathoverflow.net/questions/18898/…
Dec 9, 2010 at 14:17 answer added Mark Grant timeline score: 3
Dec 9, 2010 at 12:43 answer added Daniel Loughran timeline score: 4
Dec 9, 2010 at 12:30 history edited HKSHLZW CC BY-SA 2.5
added 123 characters in body
Dec 9, 2010 at 10:33 answer added Kostya timeline score: 7
Dec 9, 2010 at 10:33 comment added HKSHLZW Or, if essentially my assertion about surface is wrong in any way , so what i concern is just the question: can we analys the cohomology of $N$ from the one of the $M$?
Dec 9, 2010 at 10:31 comment added HKSHLZW sorry, i didn't know what the wiki is ,then i think it as a good choice . So , if adding a condition that if $N $ is oriented ,so the situation works as i concern works?
Dec 9, 2010 at 10:26 comment added José Figueroa-O'Farrill 1) why is this community wiki? Don't you expect a definite answer? 2) Are you sure about your claim concerning surfaces? $H_1(\mathbb{RP}^2) \neq H_1(S^2)$.
Dec 9, 2010 at 10:02 history asked HKSHLZW CC BY-SA 2.5