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Feb 18, 2017 at 22:25 comment added Mark Grant @wonderich: good catch. Now corrected.
Feb 18, 2017 at 22:24 history edited Mark Grant CC BY-SA 3.0
Fixed typo
Feb 18, 2017 at 20:34 comment added wonderich @Mark Grant, Actually, why do you wrote $0=i^*\alpha\in H^d(A;M)$ instead of $0=i^*\alpha\in H^d(B;M)$? Should it be trivial in the cohomology group of $B$? (Is that the same as trivial in the cohomology group of $A$?)
S Feb 18, 2017 at 20:28 history suggested miss-tery CC BY-SA 3.0
instead $\gamma\in H^d(A/B;M)$
Feb 18, 2017 at 20:19 review Suggested edits
S Feb 18, 2017 at 20:28
Feb 18, 2017 at 20:16 comment added miss-tery Thanks, +1, this is a good type of answer that I am happy to receive.
Feb 18, 2017 at 7:08 history answered Mark Grant CC BY-SA 3.0