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Jul 2, 2022 at 20:14 comment added user484289 The ongoing book of Huybrechts would also be helpful. You can find a link on his website.
Jul 1, 2022 at 16:46 vote accept Zhaoting Wei
Jul 1, 2022 at 10:01 comment added Sasha @ZhaotingWei: I added an argument to the answer.
Jul 1, 2022 at 10:01 history edited Sasha CC BY-SA 4.0
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Jun 30, 2022 at 15:40 comment added Zhaoting Wei @Sasha From Example 3.41 we can get $i^!=\mathcal{O}_X(X)[-1]\otimes i^*$ hence to prove $i^*i_*\mathcal{F}\to \mathcal{F}\to \mathcal{O}_X(-X)[2]\otimes \mathcal{F}$ is exact is equivalent to prove $\mathcal{F}\to i^!i_*\mathcal{F}\to \mathcal{O}_X(X)[-1]\otimes \mathcal{F}$ is exact. But I don't know how to prove either of them.
Jun 30, 2022 at 15:04 comment added Sasha @ZhaotingWei: What were you saying about Cor. 3.40 and Ex. 3.41 then?
Jun 30, 2022 at 13:15 comment added Zhaoting Wei @Sasha Where can I find the triangle for $i^!i_*$?
Jun 30, 2022 at 10:11 comment added crystalline Or you can just use the triangle for the unit $id \to i_*i^*$ and the "zig-zag" identities for the adjunction $(i^*, i_*)$ to get the one for the counit.
Jun 30, 2022 at 7:04 comment added Sasha I am sure you can use for $i^*$ the same argument that is used for $i^!$. Alternatively, you can use the fact that $i^!$ differs from $i^*$ by a line bundle twist and shift, and applying this to the triangle for $i^!i_*$, obtain the triangle for $i^*i_*$.
Jun 30, 2022 at 4:40 comment added Zhaoting Wei I think his Corollary 3.40 and Example 3.41 are closed to the result I want but they are still on different topic. Actually they compute $i^!$ but I am not sure if it implies the result you mentioned.
Jun 30, 2022 at 4:26 comment added Sasha I think you can find this in the book of Huybrechts (FM transforms ...).
Jun 30, 2022 at 4:22 comment added Zhaoting Wei Is it well-known? How to see it?
Jun 30, 2022 at 4:17 history answered Sasha CC BY-SA 4.0