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Jun 15, 2020 at 7:27 history edited CommunityBot
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Jun 15, 2012 at 21:17 answer added Hiraku Nakajima timeline score: 1
Jun 15, 2012 at 11:16 history edited Paul Johnson CC BY-SA 3.0
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Jun 15, 2012 at 11:11 comment added Paul Johnson Hiraku, I could have been clearer. We certainly have $\chi(F_1)=\chi(X)=\chi(F_2)$ by localization. But it appears that more is true -- namely, the betti numbers of the $F_i$ are equal. I've edited the question to make this a bit clearer.
Jun 9, 2012 at 3:22 comment added Hiraku Nakajima What do you mean by `$H^*(F_1) = H^*(F_2)$' ? There is no natural homomorphism between them. If you just compare the dimensions of RHS and LHS, they are the Euler numbers of $Y_1$ and $Y_2$, and hence are equal.
May 17, 2012 at 3:29 answer added Ben Webster timeline score: 1
May 15, 2012 at 22:11 history asked Paul Johnson CC BY-SA 3.0