Stalks (of sheaves) preserve exactness Let $\mathcal F\xrightarrow{\alpha}\mathcal G\xrightarrow{\beta}\mathcal H$ be an exact sequence of abelian sheaves on the topological  space $X$, so that $\operatorname{ker}\beta\cong \operatorname{im}\alpha$ as sheaves (in particular they  are subpresheaves of  $\mathcal  G$). Then for every point $x\in X$, also $(\operatorname{ker}\beta)_x\cong (\operatorname{im}\alpha)_x$.
We want to prove that also the sequence of abelian groups $\mathcal F_x\xrightarrow{\alpha_x}\mathcal G_x\xrightarrow{\beta_x}\mathcal H_x$ is  exact, i.e. that $\operatorname{ker}\beta_x\cong \operatorname{im}\alpha_x$; my idea would be to prove that $(\operatorname{ker}\beta)_x\cong \operatorname{ker}\beta_x$ and $ (\operatorname{im}\alpha)_x \cong \operatorname{im}\alpha_x$.
Start with $(\operatorname{ker}\beta)_x\cong \operatorname{ker}\beta_x$. Define a map $(\operatorname{ker}\beta)_x\to \operatorname{ker}\beta_x$: if $s_x\in (\operatorname{ker}\beta)_x$ is the stalk  at $x$ of $s\in \operatorname{ker}\beta(U)$, for an open set $x\in U\subseteq X$, we send it to the stalk $s_x\in \mathcal G_x$ of $s$ viewed as an element of $\mathcal G(U)$. It's obvious that such map doesn't depend on the choice of $s$ and is also  a homomorphism. Plus $\beta_x(s_x)=(\beta (s))_x$ by definition, and $\beta (s)=0$, so we constructed a homomorphism $(\operatorname{ker}\beta)_x\to \operatorname{ker}\beta_x$. Now  we need the inverse, so take $s_x\in \mathcal G_x$, with $s\in \mathcal G(U)$,  such that $\beta_x(s_x)=0$, i.e. $\beta(s)|_V=0$ for some open set $x\in V\subseteq U$. Then $s|_V\in \operatorname{ker}\beta$,  so that  we can send $s_x$ to $(s|_V)_x$. This map doesn't depend on the choice of $V$ nor on the  choice of $s$, and it is  indeed an inverse of our homomorphism, that hence is bijective.
For the other equality  the  situation is simpler: since by definition the images the stalks are  the stalks of the  images, the map $\operatorname{im}\alpha_x\to  (\operatorname{im}\alpha)_x:\alpha_x(r_x)\mapsto (\alpha(U)(r))_x$, for $r\in \mathcal F(U)$ with an open set $x\in U\subseteq X$, is  automatically a  bijective homomorphism.
At this point,  knowing that  $\operatorname{ker}\beta\cong \operatorname{im}\alpha$, we have $(\operatorname{ker}\beta)_x\cong (\operatorname{im}\alpha)_x$, and so $\operatorname{ker}\beta_x\cong \operatorname{im}\alpha_x$, meaning that passing to  the stalks preserves the  exactness of sequences. Now consider the exact sequence $0\to \mathcal F'\to \mathcal  F\to \mathcal F/\mathcal F'\to 0$, where $\mathcal  F,\mathcal  F'$ are sheaves, that yields an exact sequence $0\to \mathcal F'_x\to \mathcal  F_x\to (\mathcal F/\mathcal F')_x\to 0$, so that $(\mathcal F/\mathcal F')_x\cong \mathcal  F_x/\mathcal  F'_x$. When considering the sheafication $(\mathcal 
 F/\mathcal  F')^\dagger$, we also  have $(\mathcal 
 F/\mathcal  F')^\dagger_x\cong \mathcal  F_x/\mathcal  F'_x$. Thus I'd say that, starting from $0\to \mathcal F'\to \mathcal  F\to \mathcal F/\mathcal F'\to 0$ as above, one can only say that $0\to \mathcal F'\to \mathcal  F\to (\mathcal F/\mathcal F')^\dagger$ is exact, since  the  canonical morphism $\mathcal F/\mathcal F'\to(\mathcal F/\mathcal F')^\dagger$ is injective, while $0\to \mathcal F'_x\to \mathcal  F_x\to (\mathcal F/\mathcal F')^\dagger _x\to 0$ stays exact.
Do yo agree with my conclusions?  I'm still not familiar with sheaves so I'm a bit confused.
 A: If
$$0\to\mathcal{F}'\to\mathcal{F}\to\mathcal{F}/\mathcal{F}'\to 0$$
is exact as a sequence of presheaves, then
$$0\to\mathcal{F}'_x\to\mathcal{F}_x\to(\mathcal{F}/\mathcal{F}')_x\to 0$$
is exact as a sequence of abelian group, and since $(\mathcal{F}/\mathcal{F}')_x=(\mathcal{F}/\mathcal{F}')^\dagger_x$, and since this holds for every $x$, the sequence
$$0\to\mathcal{F}'\to\mathcal{F}\to(\mathcal{F}/\mathcal{F}')^\dagger\to 0$$
is exact as a sequence of sheaves. (In particular, $0\to\mathcal{F}'\to\mathcal{F}\to(\mathcal{F}/\mathcal{F}')^\dagger$ is exact as a sequence of presheaves, and the last map is not onto in the category of presheaves).

Note that

*

*if a sequence of presheaves $0\to A\to B\to C\to 0$ induces an exact sequence of stalks $0\to A_x\to B_x\to C_x\to 0$ for every $x$, then you cannot say that the original sequence was exact.

*however, if a sequence of sheaves induces an exact sequence of stalks for every $x$, then the original sequence of sheaves is exact.

*an exact sequence of sheaves does not induce an exact sequence of the underlying presheaves. The forgetful functor is only left exact.

