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Sep 4, 2019 at 18:48 comment added user43326 The point is that since A and B are supposed to be finite, the Kunneth isomorphism holds only in very few cases (there won't be situations like Landweber-flat), so you get Kunneth isomorphism everywhere in sight. That is either $\pi _*(E)$ is field, or one of $H_^*(A)$ or $H^*(B)$ is torsion-free. Thus $H^*(A\times B \pi_*(E))\cong H^*(A),\pi _*(E))\times H^*(B),\pi _*(E))$
Sep 4, 2019 at 16:41 comment added Monkey.D.Luffy It may not be a spectral sequence because of Kunneth formula.
Sep 4, 2019 at 13:18 comment added user43326 Actually the thread you quote says that there is natural construction for the tensor product in the case you are talking about.
Sep 3, 2019 at 21:10 review First posts
Sep 4, 2019 at 1:16
Sep 3, 2019 at 21:10 history asked Monkey.D.Luffy CC BY-SA 4.0