Is the direct image functor on sheaves of a point inclusion exact? Given the inclusion of a point $i_x:\{x\}\to X$, is the associated direct image functor $(i_x)_*: Sh(\{x\}) \to Sh(X)$ an exact functor?
I know that in general this is not the case and the direct image functor is only left exact. I also know that if $\{x\}$ is a closed subset in X, then it should be an exact functor. So I was trying to construct a counterexample by looking at the inclusion of an open subset, but to be honest, I just lack the intuition for the subject to come up with something. Any hints?
 A: I actually think, that in case of single point it will always be exact. Let
$$ 0 \to \mathcal{F} \overset{\varphi}{\to} \mathcal{G} \overset{\psi}{\to} \mathcal{H} \to 0 $$
be exact, $\mathcal{F}, \mathcal{G}, \mathcal{H} \in Sh(\{x\})$. We only need to prove, that $(i_x)_*(\psi)$ is surjective. By definition, $ (i_x)_*\mathcal{F}(U) = \mathcal{F}(\{x\}) $, if $ x \in U $, otherwise it is just $ 0 $ (because $\mathcal{F}$ is a sheaf, so on empty set it gives 0). Same thing for $\mathcal{G}, \mathcal{H}$. Next, $ (i_x)_*\psi(U) = \psi(\{x\})$ (surjective, because $\psi(\{x\}) = \psi_x$ in case of one-pointed top. space), if $x \in U $, and trivial otherwise (also surjective, because between to trivial groups). So all $(i_x)_*\psi(U)$ are surjective, then $(i_x)_*\psi$ is surjective, and the sequence is exact.
A: I gave it a go, and I think I might have solved it, but I only know the basics on sheaves so you should definetely double check my proof.
we have that $ \mathcal{F}_{x} = \mathcal{F}(\{x\}), \ \mathcal{G}_{x} = \mathcal{G}(\{x\}), \ \mathcal{H}_{x} = \mathcal{H}(\{x\}). \ $ Since $\ \ \large0 \rightarrow \mathcal{F} \rightarrow \mathcal{G} \rightarrow \mathcal{H} \rightarrow \large0 \ $ is exact, also $ \ \large0 \rightarrow \mathcal{F}_{x} \rightarrow \mathcal{G}_{x} \rightarrow \mathcal{H}_{x} \rightarrow \large0 \ $ is exact.
In order to prove that $ \ \ \large0 \rightarrow i_{\star}(\mathcal{F}) \rightarrow i_{\star}(\mathcal{G}) \rightarrow i_{\star}(\mathcal{H}) \rightarrow \large0 \ $ is exact, we can just prove that it is exact on the stalks. But this is indeed the case:
Let $y \in X \ \ \text{s.t.  there's an open set} \ U \subset X \ \text{with } x \notin U.$ Then $i_{\star}(\mathcal{F})_{y} = i_{\star}(\mathcal{G})_{y} = i_{\star}(\mathcal{H})_{y} = \large0$
If $ \ \  \forall \ \  U \subset X \text{ such that } y \in U \text{ we also have } x \in U, \ $ then $ \ i_{\star}(\mathcal{F})_{y} = \mathcal{F(\{x\})}, \ i_{\star}(\mathcal{G})_{y} = \mathcal{G(\{x\})}, \ i_{\star}(\mathcal{H})_{y} = \mathcal{H(\{x\})}$.
In both cases the sequence $ \ \ \large0 \rightarrow i_{\star}(\mathcal{F})_{y} \rightarrow i_{\star}(\mathcal{G})_{y} \rightarrow i_{\star}(\mathcal{H})_{y} \rightarrow \large0 \ $ is exact. Therefore $ \ \ \large0 \rightarrow i_{\star}(\mathcal{F}) \rightarrow i_{\star}(\mathcal{G}) \rightarrow i_{\star}(\mathcal{H}) \rightarrow \large0 $ is also exact.
