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Jan 9, 2021 at 3:41 comment added user267839 hhhm the only unsolved problem stays why the lecturer who gave the talk where this subfunctor was introduced, remarked shortly that it is well defined by "semicontinuity".
Jan 9, 2021 at 3:03 comment added user267839 @Sasha: Yes, you suggestion is reasonable. Then with notations from above for $f: S \to U$ and $\hat{f}$ by flat base change formula we obtain $R^1 (\pi_S)_* \bar{f}^* \mathcal{F}= f^* R^1(\pi_U)_* \mathcal{F} =0$ since $\pi_U: X \times U \to U$ and $\pi_S$ are flat und thus the semicontinuity theorem isn't neccessary here.
Jan 9, 2021 at 3:02 history edited user267839 CC BY-SA 4.0
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Jan 8, 2021 at 4:54 comment added Sasha It makes more sense to replace the condition $H^1 = 0$ by the ondition $R^1\pi_* = 0$, where $\pi$ is the projection to $U$. This condition is equivalent to the $H^1$ condition when $U$ is affine and it is preserved by base change.
Jan 7, 2021 at 23:34 history edited user267839 CC BY-SA 4.0
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Jan 7, 2021 at 23:28 history edited user267839 CC BY-SA 4.0
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Jan 7, 2021 at 22:59 history edited user267839 CC BY-SA 4.0
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Jan 7, 2021 at 22:53 history asked user267839 CC BY-SA 4.0