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May 31, 2011 at 15:40 vote accept th.ng
May 31, 2011 at 15:29 comment added algori .. However, it is true, as Dan says in his answer, that if the sheaf $R^{q_0}f_! \phi$ is constant, then everything works fine.
May 31, 2011 at 15:26 comment added algori Here is a small remark: this statement is false without additional assumptions. Take e.g. the map $\mathbb{A}^1\setminus\{0\}\sqcup \{0\}\to \mathbb{A}^1$. All closed fibers are points, but the cohomology of the target is not the same as the cohomology of the source.
May 31, 2011 at 15:01 answer added Dan Petersen timeline score: 3
May 31, 2011 at 14:53 history edited th.ng CC BY-SA 3.0
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May 31, 2011 at 14:40 history asked th.ng CC BY-SA 3.0