Timeline for Is the perfection (perfect closure) presheaf a sheaf?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| S Mar 17, 2020 at 23:21 | history | suggested | RobPratt |
Added a tag
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| Mar 17, 2020 at 22:54 | review | Suggested edits | |||
| S Mar 17, 2020 at 23:21 | |||||
| Mar 17, 2020 at 22:43 | answer | added | Johan | timeline score: 4 | |
| Mar 17, 2020 at 11:01 | comment | added | A.G | It was obvious. Sorry, and thank you. | |
| Mar 17, 2020 at 10:48 | comment | added | Piotr Achinger | Let $(A_i)$ be an infinite sequence of rings such that $(\prod A_i)_{\rm pf} \neq \prod (A_i)_{\rm pf}$, and let $X$ be the disjoint union of $U_i = \operatorname{Spec} A_i$. Note that $X$ is not affine since it is not quasi-compact. Still, $\Gamma(X, \mathcal{O}_X) = \prod A_i$, so $\Gamma(X, \mathcal{O}_X)_{\rm pf} = (\prod A_i)_{\rm pf}$. Consider sheaf condition the covering of $X$ by $U_i$, it simply says $\mathcal{F}(X) = \prod \mathcal{F}(U_i)$, which is not true for your presheaf. | |
| Mar 17, 2020 at 10:32 | comment | added | A.G | @Piotr Achinger Why that example shows that it is not a sheaf? Is it something easy that I have to think more? (sorry if it is). | |
| Mar 16, 2020 at 20:51 | comment | added | Piotr Achinger | I think YCor's example in the linked question shows that this is not quite a sheaf unless you disallow infinite coverings. However, if you sheafify your presheaf, then it will still be true for affine (or quasi-compact) $U$ that sections over $U$ of that sheaf will equal $\Gamma(U, \mathcal{O}_X)_{\rm pf}$. | |
| Mar 16, 2020 at 19:38 | history | edited | A.G | CC BY-SA 4.0 |
added 26 characters in body
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| Mar 16, 2020 at 19:25 | history | asked | A.G | CC BY-SA 4.0 |