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?

3$\begingroup$ Maybe an example where sheafification is needed is what you need? $\endgroup$ – Mariano SuárezÁlvarez Mar 23 '10 at 16:56
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.

1$\begingroup$ Ah, I see! One applies implicitly the (right adjoint) inclusion $Shv\to PShv$ which does NOT commute with colimits. Thanks, Zoran. $\endgroup$ – roger123 Mar 24 '10 at 10:03
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 $zc$ doesn't vanish with order two at $O$.

$\begingroup$ I think the question really is: Why is it not a sheaf if sheafification commutes with colimits? $\endgroup$ – Andrea Ferretti Mar 23 '10 at 18:13

6$\begingroup$ 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. $\endgroup$ – Zoran Skoda Mar 23 '10 at 18:20

1$\begingroup$ 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. $\endgroup$ – Andrea Ferretti Mar 24 '10 at 1:23
I don't really know what you need, but here is my understanding:
There are two ways to define a sheaf of abelian groups:
I. as a local homeomorphic projection with some properties on fibers.
II. as a presheaf (with S1. gluing and S2. localglobal uniqueness properties).
Sheafification is a functor from II to I, which can be viewed as a natural transformation of sheaf functors. For a presheaf satisfying S1 and S2, sheafification is an equivalence. Thus one may want to identify I and II as a definition of a sheaf, but this is NOT the correct thing to do, and the obstruction to identifying them is exactly the cohomology of the subsheaf.
In detail, on one hand, the group structures in I are given fiberwise and homomorphisms of sheaves are continuous maps which are homomorphisms on the fibers; on the other hand, the group structures in II are given more globally on open sets. So passing from I to II is exactly asking the lifting question from local to global.
When we define exact sequences stalkwise, it may fail to be surjective when passing to global, for example, let X be $C^*$, the exact sequence $0 \rightarrow Z \rightarrow \mathcal{O} \rightarrow \mathcal{O^*} \rightarrow0 $, which fails to be surjective since log is not defined on $C^*$.
That means we have some global sections in the "quotient sheaf" which are not quotients of global sections; this is because some local data do glue together in the locally compatible sense, but after gluing fail to be quotients of global sections. See here for another example.
This suggests that it may fail if we want to define a quotient sheaf in the brutal way that you suggested, because you may fail to have enough global data to realize a gluing. For a sheaf functor you really mean a presheaf functor, despite it commuting with colimit, you also need to check S1 & S2 for being a sheaf.
It is worth mentioning there are also important cases of presheaves which fail S1, so we have to do sheafification in order to give sheaf cohomology. That is the presheaf of cochains. In this case we lose the clarity of supports and restriction, and this give rise to the difficulty of considering cohomology of subsets and of pairs.