Timeline for Cohomology of the pullback of a coherent sheaf (considered as an abelian sheaf)
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 22, 2019 at 18:43 | comment | added | cgodfrey | @schematic_boi it's definitely not. I was confused and off base. My bad all! | |
| Apr 21, 2019 at 18:28 | comment | added | user138661 | @cgodfrey there is nothing stopping you, that is true, but the question asks about cases where the cohomology is *non-zero$, not zero. How is your comment relevant to the question? | |
| Apr 14, 2019 at 14:24 | comment | added | S. carmeli | @StepanBanach Ohhh I see! I switched the role of the spaces (I had in mind that $X$ was non-affine but you written the opposite). Ill think about it this question seems to make much more sense when you read it appropriately. | |
| Apr 14, 2019 at 13:42 | comment | added | user137767 | @S.carmeli thak your for your comment. We start with a quasi-coherent sheaf on a possibly non-affine (but separated) scheme $Y$, pull it back to a possibly non-quasi-coherent sheaf on $X$. It is true that the cohomology is not changed by pull-back. It is true that the cohomology of a quasi-coherent sheaf on $X$ vanishes. How does it follow that the cohomology of $f^{-1}(F)$, which is the same as the cohomology of $F$, vanishes? | |
| Apr 14, 2019 at 13:36 | comment | added | S. carmeli | @StepanBanach The fact that the cohomology is unchanged by $f^{-1}$ together with the vanishing of the cohomology of quasi-coherent sheaves on an affine variety together shows that there is no such example, $H^1(X,f^{-1}(F))$ always vanishes in this case. The fact that $f^{-1}(F)$ is not quasi-coherent has nothing to do with it. | |
| Apr 14, 2019 at 9:55 | comment | added | user137767 | @cgodfrey $f^{-1}(F)$ is not necessarily a quasi-coherent sheaf (the quasi-coherent sheaf is $f^{-1}(F)\otimes_{f^{-1}(O_Y)}O_X$). The question explicitly states that we are taking the pullback of an abelian sheaf. The cohomology of an arbitrary abelian sheaf on an affine Noetherian scheme does not necessarily vanish. | |
| Apr 14, 2019 at 9:50 | comment | added | user137767 | @S.carmeli the question asked for an example where the cohomology of $f^{-1}(F)$ does not vanish. Do you have such an example? | |
| Apr 14, 2019 at 5:40 | comment | added | cgodfrey | What's more relevant than the forgetful functor from schemes to spaces not being full is that the forgetful functor from quasi-coherent sheaves to sheaves of abelian groups is exact. So you can compute cohomology of a quasi-coherent sheaf $\mathcal{F}$ as the cohomology of the underlying sheaf of abelian groups. | |
| Apr 14, 2019 at 5:36 | comment | added | cgodfrey | I don't see anything stopping us from taking $Y = X$ and $f = \mathrm{id}$. Then Certainly $H^1(X, f^{-1}\mathcal{F}) = H^1(X, \mathcal{F}) = 0$ since $\mathcal{F}$ is a quasi-coherent sheaf on an affine scheme. | |
| Apr 13, 2019 at 22:41 | comment | added | S. carmeli | I do not see the relevance of the "may be confusing" part here. If $f:X\to Y$ is a homeomorphism then $H^1(Y,F)\cong H^1(X,f^{-1}(F))$ for every sheaf $F$, coherent or not. | |
| Apr 13, 2019 at 22:29 | history | asked | user137767 | CC BY-SA 4.0 |