Sheafification, stalks and quotient I gave a problem that I can't finish by myself. Any help would be appreciated.
Consider a sheaf $\mathcal{F}$ of abelian groups on a topological space.
I would like to show that given two sheaves $\mathcal{F}, \mathcal{F}'$ we get $\mathcal{F}_x/\mathcal{F}'_x \cong (\mathcal{F}/\mathcal{F}')_x$. This question has been addressed here: Stalk of the quotient presheaf, but with no accepted answer, and the answers that are there didn't help me much. I was hoping that this could be proven without going into category theory explicitly (as was suggested in one answer to that question). I already figured that one should take the sheaf generated by $\mathcal{F}/\mathcal{F}'$, call it $\mathcal{G}$, and then one can consider the following exact sequence: $\mathcal{F} \rightarrow \mathcal{F}' \rightarrow \mathcal{G}$, where the second map is the composition of the quotient map with the natural morphism $\tau: \mathcal{F}/\mathcal{F}' \rightarrow \mathcal{G}$. Where to go from here, or if I am right so far, I do not know. Thanks in advance.
 A: You can approach this directly. It seems to me that you are using $\mathcal{F}/\mathcal{F}'$ to refer to the pre-sheaf quotient, I feel I should mention that it is more standard to give the pre-sheaf some other notation (such as $\mathcal{Q}$) and then use $\mathcal{F}/\mathcal{F}'$ for the sheafification of the pre-sheaf quotient, to emphasise that this sheaf is a cokernel for $\psi:\mathcal{F}'\rightarrow \mathcal{F}$. This is the notation that I will use.
An element of the stalk $\mathcal{Q}_x$ is a pair $(s,U)$ such that $s \in \mathcal{Q}(U)$, i.e. $s = \bar{f}$ where $f \in \mathcal{F}(U)$ and $\bar{f}$ is the image of $f$ in the quotient $\mathcal{F}(U)/(\psi(U)(\mathcal{F}'(U))$.
Then we may define a group homomorphism $\pi_x:\mathcal{F}_x\rightarrow \mathcal{Q}_x$ by $(f,U)\mapsto (\bar{f},U)$. Then you need to show that this is surjective and that the kernel is $\psi_x(\mathcal{F}'_x)$.
If we take the natural projection homomorphism $\pi:\mathcal{F}\rightarrow \mathcal{Q}$, $\pi_x$ is just the map of stalks induced by $\pi$, so it really is a very natural map to use. In fact, this approach is exactly how you show that the sequence of sheaves $\mathcal{F}'\rightarrow\mathcal{F}\rightarrow \mathcal{F}/\mathcal{F}'\rightarrow 0$ is exact, since $\mathcal{F}\rightarrow \mathcal{F}/\mathcal{F}'$ is the composition of $\pi$ and the canonical map $\varphi:\mathcal{Q}\rightarrow \mathcal{F}/\mathcal{F}'$, and we know that $\varphi_x$ is an isomorphism on stalks for all $x$.
A: We have an exact sequence
$$0 \to \mathcal F' \to \mathcal F \to \mathcal F/\mathcal F' \to 0$$
This means that this sequence is exact at stalks, hence for all $x$ we have an exact seuqnece
$$0 \to \mathcal F'_x \to \mathcal F_x \to (\mathcal F/\mathcal F')_x \to 0$$
which immediately shows $(\mathcal F/\mathcal F')_x \cong \mathcal F'_x / \mathcal F_x$.
