Consider a sheaf of abelian groupS $\mathscr F$ over a topological space $X$. If $\mathscr F'$ is a subsheaf of $\mathscr F$ (over $X$), then we can construct the quotient presheaf $\mathscr F/\mathscr F'$ in the following way:

$$(\mathscr F/\mathscr F')(U):=\mathscr F(U)/\mathscr F'(U)$$

Now I don't understand why it is true that $(\mathscr F/\mathscr F')_x=\mathscr F_x/\mathscr F'_x$ for every $x\in X$.

Notation: $\mathscr F$ is a presheaf (of abelian groups) over $X$ and $x\in X$, with the notation $\mathscr F_x$ I mean the stalk at $x$.


This follows from the fact that direct limits are exact in the category of (direct systems of) abelian groups (note: this is not true of arbitrary colimits). I suggest you try to prove this to yourself at least once as it's an important exercise.

  • $\begingroup$ I agree, this is the way to approach the problem. I might add that, since you are working on the level of presheaves, you need the added fact that sheafification preserves these colimits (this is essentially what was pointed out below). This is, of course, if you want to get the result for the corresponding sheafification of the quotient. $\endgroup$ – user113529 Feb 16 '14 at 17:34

$\def\cF{\mathcal{F}}$ This follows from the following very nice fact:

If $\cF$ is a presheaf on $X$ and $\widetilde{\cF}$ is its sheafification, then for all $x \in X$, the natural map $\cF_x \to \widetilde{\cF}_x$ is an isomorphism.

This follows from the explicit construction of the sheafification as sections of the disjoint union of stalks of the presheaf. I'm not sure if there's an easier way to see it.


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