Timeline for Does Grothendieck duality hold without taking RHom?
Current License: CC BY-SA 4.0
6 events
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| Feb 21, 2022 at 23:52 | comment | added | Adam | I edited the question, I hope it is clearer now? Basically the context is not that important, I'm interested in the usual situations when Grothendieck Duality holds as a quasi-- isomorphism of complexes of sheaves and I am wondering if one needs to go to the full RHom for this quasi--isomorphism to be true (Hom always means sheaf-Hom here) or if it is true also for $H^0RHom$ which is the $Hom$ in the derived category? I apologise if I'm missing something and my question is badly posed, please help enlighten me in that case. Thank you! | |
| Feb 21, 2022 at 23:46 | history | edited | Adam | CC BY-SA 4.0 |
Edited question to incorporate comments
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| Feb 18, 2022 at 11:14 | comment | added | Leo Alonso | The question is not clear to me. Does $\mathcal{S}(X)$ refers to $\mathcal{O}_X-Mod$ or $Qco(X)$? There is no trace for general modules. I suggest lookin at Lipman's Lecture Notes 1960 (also at math.purdue.edu/~jlipman/Duality.pdf) to grasp the missing details. | |
| Feb 18, 2022 at 7:33 | comment | added | R. van Dobben de Bruyn | Isn't $\mathbf{Hom}(A,f^!A')$ an $\mathcal O_X(X)$-module instead of a sheaf of $\mathcal O_X$-modules? What does it mean to take $Rf_*$ of such a thing? | |
| Feb 18, 2022 at 5:28 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Embedded hyperlink to former question.
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| Feb 17, 2022 at 18:57 | history | asked | Adam | CC BY-SA 4.0 |