## Why must one sheafify quotients of sheaves?

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Let $\mathcal{F}$ and $\mathcal{G}$ be two sheaves (of abelian groups) on a topological space $X$ such that $\mathcal{G}(U)$ is a subgroup of $\mathcal{F}(U)$ for every open set $U$ in $X$. The sheaf associated to the presheaf $P(\mathcal{F}/\mathcal{G})$ defined by ```$$
U\mapsto \mathcal{F}(U)/\mathcal{G}(U)
$$``` is called the quotient sheaf $\mathcal{F}/\mathcal{G}$. The associated sheaf functor is left adjoint to the inclusion functor, so it commutes with colimits and in particular with quotients. My question is: Why must one sheafify the presheaf $P(\mathcal{F}/\mathcal{G})$ then?

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Maybe an example where sheafification is needed is what you need? – Mariano Suárez-Alvarez Mar 23 2010 at 16:56

## 2 Answers

For presheaves (of sets or groups) we know what this particular (or any) colimit operation is: apply the operation objectiwise (for each $U$). Now the sheafification preserves colimits, hence we apply sheafification to a colimit cocone on presheaves to obtain a colimit cocone in sheaves. Doing sheafification to the presheaves which are alerady sheaves does nothing to them, but it, by the right exactness, does the correct thing to the colimit. This proves that the sheafification following the colimit in presheaves is the correct way to compute the colimit, and in that we did use the right exactness of the sheafification essentially. The fact that it is necessary does not follow from the general nonsense as there are both examples where we accidentally do not need a sheafification step and those where we do need. For the limit constructions on sheaves we never need the sheafification because the embedding of the sheaves into presheaves is left exact hence we can simply compute the limits in presheaves. It seems you had somehow an opposite impression.

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Ah, I see! One applies implicitly the (right adjoint) inclusion $Shv\to PShv$ which does NOT commute with colimits. Thanks, Zoran. – roger123 Mar 24 2010 at 10:03

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You ask: "Why must one sheafify the presheaf $P(\mathcal{F}/\mathcal{G})$ then?". The answer is: "Because it is not a sheaf!" Here is an example.

Let $X=\mathbb P^1_k, \mathcal F = \mathcal O, \mathcal G=\mathcal O (-2.O)$, where $O$ denotes the origin $O=(0:1)$ . Define $\infty=(1:0)$ and let $z$ be the coordinate on $\mathbb P^1\setminus \infty$ .

Consider the covering of X by the open subsets $U_0=X\setminus \infty$ and $U_\infty =X\setminus O$. Let me denote the presheaf $P(\mathcal{F}/\mathcal{G})$ just by $P$.

Then $class(z)\in P(U_0)$ and $class(0)\in P(U_\infty)$ are sections of $P$ over the two open sets $U_0$ and $U_\infty$ of our covering which coincide on their intersection $U_{0\infty}$ for the excellent reason that $P(U_{0\infty})=0$ ! [Actually $P(U)=0$ for any open subset $U\subset X$ not containing $O$]

But these compatible sections cannot be glued to a global section of $P$ on $X$. Indeed a section of $P$ on $X$ is just a constant $c\in k$ , since $\mathcal O(X)=k$ and $\mathcal O(-2.O)(X)=0$. But that constant $c$ cannot be the glued global section , because its restriction to $P(U_0)$ is $class(c)$ and in $P(U_0)$ we have $class(c) \neq class(z)$ since $z-c$ doesn't vanish with order two at $O$.

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I think the question really is: Why is it not a sheaf if sheafification commutes with colimits? – Andrea Ferretti Mar 23 2010 at 18:13

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Commutes means that sheafification of a (colimit in PShv) is (colimit in Shv) of a sheafification (the latter step is empty operation on sheaves of course) and NOT that sheafification of a colimit in PShv is just a colimit in PShv. – Zoran Škoda Mar 23 2010 at 18:20
Yes, I know, I was just trying to point out what the original problem was. I guess roger was aware of examples of quotient presheaves which are not sheaves. – Andrea Ferretti Mar 24 2010 at 1:23
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