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Aug 31, 2016 at 22:29 comment added jdc @AlexDegtyarev: Would you happen to have a reference handy?
Jun 10, 2015 at 8:31 vote accept Shiquan Ren
Jan 14, 2015 at 9:54 answer added Dan Petersen timeline score: 5
Jan 14, 2015 at 9:53 comment added Alex Degtyarev There is such a version, and it is compatible with the products (in the usual sense: differentials obey the Leibnitz rule and all isomorphisms involved are multiplicative). Of course, you cannot get the product structure of the limit term, just the corresponding graded ring.
Jan 14, 2015 at 9:20 comment added user43326 Doesn't it suffice to simply dualize? For the cup product structure, you need more than $H^*(X,k)$ and $H^*(\pi , k)$. You will need $H^*(\pi , H^*(X;k))$ (note that $\pi $ usually acts non-trivially on $H^*(X;k)$).
Jan 14, 2015 at 9:15 history edited Shiquan Ren CC BY-SA 3.0
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Jan 14, 2015 at 9:02 history asked Shiquan Ren CC BY-SA 3.0