Timeline for Cohomology classes annihilated by pullbacks
Current License: CC BY-SA 2.5
3 events
| when toggle format | what | by | license | comment | |
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| Mar 7, 2010 at 22:01 | comment | added | algori | Somnath -- yes, absolutely, there should be some assumptions; either $\pi_1(B)=e$ or $\pi_1(G)=e$ would do, I think. | |
| Mar 7, 2010 at 21:34 | comment | added | Somnath Basu | For $(2)$ for any manifold $B$ don't you need either $H^2(B;\mathbb{Z})$ to be zero or $G$ simply connected to get a trivial bundle. For example, take $B=S^1\times S^2$ with total space $M=S^1\times S^3$ and $\pi=\textrm{id}\times\textrm{Hopf}$. Then the required map is not injective on $H^3$'s. | |
| Mar 7, 2010 at 20:39 | history | answered | algori | CC BY-SA 2.5 |