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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Dec 1, 2016 at 16:48 comment added Daniel Miller The lower shriek is this case is just extension by zero, so $\Gamma(W, i_! \mathcal{O}_U) = \Gamma(W,\mathcal{O}_U)$ if $W\subset U$, and $0$ otherwise. So in this case $\Gamma(i_! \mathcal{O}_U) = \Gamma(i_! \mathcal{O}_V) = 0$, since $U$ and $V$ are strict subsets of $X$.
Dec 1, 2016 at 13:49 comment added user113988 Sorry for obvious question. Why is $\Gamma(i_! \mathcal{O}_U) \oplus \Gamma(i_! \mathcal{O}_V) = 0$?
Mar 17, 2015 at 19:29 vote accept Zhaoting Wei
Mar 17, 2015 at 17:49 history answered Daniel Miller CC BY-SA 3.0