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Quotient and tensor sheaves are defined by sheafification which is somehow implicit. Is there any easy way to handle these constructions?

Could we ignore the sheafification when we are working in affine open subsets?

For example, let $X$ be a scheme and $\mathcal F$ and $\mathcal G$ be quasicoherent sheaves. Let $U\subseteq X$ be an affine open. Is it true that $(\mathcal F/\mathcal G)(U) = \mathcal F(U)/\mathcal G(U)$ and $\mathcal F\otimes_{\mathcal O_X}\mathcal G(U) = \mathcal F(U)\otimes_{\mathcal O_X(U)}\mathcal G(U)$?

How about other constructions involving sheafification? (for example, image, cokernel, ann, nil and so on)

Thank you in advance.

Hydrogen
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    Great question about something that I still find confusing to this day. – Tabes Bridges Dec 11 '20 at 19:02
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    I tend to just pretend it is what I hope it is, and then pray I can work locally on the stalks where it is what I hope it is. – Mummy the turkey Dec 11 '20 at 19:06
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    That is a general phenomenon that : constructions involving quasicoherent sheaves that involve sheafification for general sheaves don’t require sheafification when considered on the distinguished affine base. This lies in the fact that many operation commute with localization, therefore the presheaf is already a sheaf on the distinguished open base. – yi li Jan 30 '23 at 02:37

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The key statement to verify is whether the thing you're looking at commutes with localization. If it does, then it commutes with sections on affine opens: start by defining the presheaf which takes the naive value on every open set, and then note that on principal affine opens $D(f)$ you get exactly the value you would expect from the quasi-coherent sheaf associated to the operation on the module of global sections. As the principal affine opens form a basis for the topology on an affine scheme and a sheaf can be specified by its values on a basis, the result follows.

In particular, this means that kernels, cokernels, images, and any other "homological" constructions commute with the associate sheaf functor on an affine scheme because localization is exact. Tensor products and nilradicals commute with localization, so they commute with the associated sheaf functor, too. Annihilators don't always commute with localization, so you need to be more careful with them.

Let me just mention one alternate neat proof that tensor products commute with sheafification, which I originally saw at the Stacks Project: consider the morphism of ringed spaces $\pi:(\operatorname{Spec} R,\mathcal{O}_{\operatorname{Spec} R})\to (\{pt\},R)$. Then the associated sheaf functor is $\pi^*$, which means it commutes with tensor products by the standard proof.

KReiser
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  • What does that mean by 'commutes with localization'? And why on $D(f)$ we get exactly the expected value? – Hydrogen Dec 13 '20 at 04:56
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    Well, take quotients for example: if $R$ is a ring, $S\subset R$ is a multiplicative subset, and $A$ is an $R$-submodule of an $R$-module $B$, then $S^{-1}(B/A)\cong (S^{-1}B)/(S^{-1}A)$ naturally. "Commutes with localization" means the localization of whatever thing you're doing is the same as doing whatever you're doing to the localization. Secondly, if $\mathcal{F}$ is a quasicoherent sheaf on an affine scheme $X$, then $\mathcal{F}(D(f))=\mathcal{F}(X)_f$, then one applies the fact that you're doing something that commutes with localization. – KReiser Dec 13 '20 at 05:19
  • Maybe I do not understand it well. Take quotient as an example. Let $U$ be an affine open set. I think your idea is to show $\mathcal F(D(f))/\mathcal G(D(f)) = (\mathcal F/\mathcal G)(U)_f$ and then use the gluing uniqueness. But from 'commutes with localization', we only know $\mathcal F(D(f))/\mathcal G(D(f)) = \mathcal F(U)_f/\mathcal G(U)_f$. How to show $\mathcal F(U)_f/\mathcal G(U)_f = (\mathcal F/\mathcal G)(U)_f$? – Hydrogen Dec 13 '20 at 05:50
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    Say $U=\operatorname{Spec} A$. Then $\mathcal{F}(U)_f/\mathcal{G}(U)_f=(\mathcal{F}(U)/\mathcal{G}(U))_f$, which is exactly the value on $D(f)$ of the sheaf associated to the $A$-module $\mathcal{F}(U)/\mathcal{G}(U)$. – KReiser Dec 13 '20 at 06:21
  • Sorry, but I'm still confused. I think we should show $(\mathcal F/\mathcal G)(U)_f = [\mathcal F(U)/\mathcal G(U)]_f$. But 'commutes with localization' only tells us $[\mathcal F(U)/\mathcal G(U)]_f = \mathcal F(U)_f/\mathcal G(U)_f$. – Hydrogen Dec 13 '20 at 18:48
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    Take the sheaf on $U$ associated to that $A$-module $\mathcal{F}(U)/\mathcal{G}(U)$. The sections of this on $U$ are $\mathcal{F}(U)/\mathcal{G}(U)$, and the sections on $D(f)$ are $[\mathcal{F}(U)/\mathcal{G}(U)]_f$, which is exactly the sections of the quotient presheaf $\mathcal{F}/\mathcal{G}$. As principal affine opens form a basis and this specifies a sheaf on that basis, we're done. – KReiser Dec 13 '20 at 20:32
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    Alternately, look at the canonical exact sequence $0\to\mathcal{F}\to\mathcal{G}\to\mathcal{F}/\mathcal{G}\to 0$: as sections over $U$ is an exact equivalence between quasicoherent sheaves on $U$ and $A$-modules, this corresponds to the exact sequence $0\to \mathcal{F}(U)\to\mathcal{G}(U)\to \mathcal{F}/\mathcal{G}\to 0$, which is isomorphic to $0\to\mathcal{F}(U)\to\mathcal{G}(U)\to \mathcal{F}(U)/\mathcal{G}(U)\to 0$, so taking associated sheaves we see that the third item is both $\mathcal{F}/\mathcal{G}$ and the sheaf associated to $\mathcal{F}(U)/\mathcal{G}(U)$. – KReiser Dec 13 '20 at 20:34
  • Okay. Now I understand that, as presheaves, $[\mathcal F(U)/\mathcal G(U)]^\sim(D(f)) = (\mathcal F/\mathcal G)^{pre}(D(f))$. But how to conclude they are actually equal as sheaves? – Hydrogen Dec 13 '20 at 23:45
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    They might not be the same, but sheafification won't alter the sections on the principal affine opens: any section of the quotient sheaf on a principal affine open is glued from sections on an open cover, which by refinement we may take to be by smaller principal affine opens, and that already satisfies the sheaf condition. The proof that locality is satisfied is the same. – KReiser Dec 14 '20 at 00:00
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    Ohhhh! I think I got your idea. Do you mean the following statement? Let $\mathcal F$ be a sheaf and $\mathcal G$ be a presheaf and $\mathcal B$ be a topological basis of $X$. If $\mathcal F(U) = \mathcal G(U)$ for all $U\in \mathcal B$, then $\mathcal F = \mathcal G^\dagger$. – Hydrogen Dec 14 '20 at 04:39
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    Yes, that's correct. – KReiser Dec 14 '20 at 05:55