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Apr 4, 2012 at 13:22 vote accept KotelKanim
Apr 4, 2012 at 13:22 comment added KotelKanim Oops... that was silly. Thanks for the answer.
Apr 4, 2012 at 13:20 history edited Neil Strickland CC BY-SA 3.0
Argument spelled out in more detail.
Apr 4, 2012 at 13:05 comment added Neil Strickland Cohomology is contravariant.
Apr 4, 2012 at 12:52 comment added KotelKanim I don't see how the last line gives a contradiction. You have a non-zero element in $H^3(K)$, which is mapped to zero in $H^3(Y)$, what's wrong with that?
Apr 4, 2012 at 12:14 history answered Neil Strickland CC BY-SA 3.0