$\text{Hom}_{\mathcal{O}_X}(\cdot, \mathcal{G})$ is an exact functor

Question

Suppose that $(X, \mathcal{O}_X)$ is a ring space. If $\mathcal{F}, \mathcal{G}$ are sheaves of $\mathcal{O}_X$-modules then the assignment $$U \mapsto \text{Hom}_{\mathcal{O}_X|U}(\mathcal{F}|U, \mathcal{G}|U)$$ makes a sheaf of $\mathcal{O}_X$-modules. (The restriction should be the natural one.)

The question is to show that the functor $\text{Hom}_{\mathcal{O}_X}(\cdot, \mathcal{G})$ is left-exact (for fixed sheaf $\mathcal{G}$) i.e. if we have an exact sequence of sheaves $$\mathcal{F}' \rightarrow \mathcal{F} \rightarrow \mathcal{F}'' \rightarrow 0$$ then the sequence $$0 \rightarrow \text{Hom}_{\mathcal{O}_X}(\mathcal{F}'', \mathcal{G}) \rightarrow \text{Hom}_{\mathcal{O}_X}(\mathcal{F}, \mathcal{G}) \rightarrow \text{Hom}_{\mathcal{O}_X}(\mathcal{F}', \mathcal{G})$$ is exact.

My attempt is as follow: by definition of exact-sequence-of-sheaves, we have to show that the induced sequence on stalks $$0 \rightarrow (\text{Hom}_{\mathcal{O}_X}(\mathcal{F}'', \mathcal{G}))_x \rightarrow (\text{Hom}_{\mathcal{O}_X}(\mathcal{F}, \mathcal{G}))_x \rightarrow (\text{Hom}_{\mathcal{O}_X}(\mathcal{F}', \mathcal{G}))_x$$ is exact. Let's consider the first one where we need to show the map is injective. Let's take two germs $f, g \in (\text{Hom}_{\mathcal{O}_X}(\mathcal{F}'', \mathcal{G}))_x$ such that their images are equals in $(\text{Hom}_{\mathcal{O}_X}(\mathcal{F}, \mathcal{G}))_x$. Let denote the map $ \mathcal{F} \rightarrow \mathcal{F}''$ by $\delta$. Then we by definition of the stalk (by direct limit), we have $$(f \circ \delta)_W = (g \circ \delta)_W \text{ i.e. } f_W \circ \delta_W = g_W \circ \delta_W$$ (as sheaf hom) for some open set $W$ containing $x$; and we want to show that $f_Z = g_Z$ for some open set $Z$ containing $x$. But this does not seems possible since we only know that $\delta_W$ is surjective under the limit. So I don't find anyway to produce such set $Z$.

EDIT: Lemma 16.3 in this document is probably what I need. But unfortunately, the proof is omitted. Alternatively, it was also stated without proof that

whether or not a morphism of sheaves is a monomorphism, epimorphism, or isomorphism can be tested on the stalks

on Wikipedia.

Answer 1

By definition, a sequence $\mathcal{F}^{\prime}\to\mathcal{F}\to\mathcal{F}^{\prime\prime}\to0$ of $\mathcal{O}_X$-modules is right-exact if and only if \begin{equation} \forall x\in X,\,\mathcal{F}^{\prime}_x\to\mathcal{F}_x\to\mathcal{F}^{\prime\prime}_x\to0\,\,(*) \end{equation} are right-exact sequences of $\mathcal{O}_{X,x}$-modules.

Applying $\hom_{\mathcal{O}_X}(\_,\mathcal{G})$, at stalks level one has: \begin{equation} \forall x\in X,\,0\to\hom_{\mathcal{O}_{X,x}}(\mathcal{F}^{\prime\prime}_x,\mathcal{G}_x)\to\hom_{\mathcal{O}_{X,x}}(\mathcal{F}_x,\mathcal{G}_x)\to\hom_{\mathcal{O}_{X,x}}(\mathcal{F}^{\prime}_x,\mathcal{G}_x)\,\,(**) \end{equation} because the (contravariant) functors $\hom_{\mathcal{O}_{X,x}}(\_,\mathcal{G}_x)$ are right-exact, the sequences $(**)$ are left-exact.

By $(*)$ one can affirm that \begin{equation} \forall U\subseteq X\,\text{open},\,\mathcal{F}^{\prime}_{|U}\to\mathcal{F}_{|U}\to\mathcal{F}^{\prime\prime}_{|U}\to0\,\,(\sharp) \end{equation} are right-exact sequences of $\mathcal{O}_{X|U}$-modules; applying the (contravariant) right exact functors $\hom_{\mathcal{O}_{X|U}}(\_,\mathcal{G}_{|U})$ to $(\sharp)$ one has the following left-exact sequences of $\mathcal{O}_{X|U}$-modules: \begin{gather} \forall U\subseteq X\,\text{open},\,0\to\hom_{\mathcal{O}_{X|U}}(\mathcal{F}^{\prime\prime}_{|U},\mathcal{G}_{|U})\to\hom_{\mathcal{O}_{X|U}}(\mathcal{F}_{|U},\mathcal{G}_{|U})\to\hom_{\mathcal{O}_{X|U}}(\mathcal{F}^{\prime\prime}_{|U},\mathcal{G}_{|U})\\ 0\to(\hom_{\mathcal{O}_X}(\mathcal{F}^{\prime\prime},\mathcal{G}))(U)\to(\hom_{\mathcal{O}_X}(\mathcal{F},\mathcal{G}))(U)\to(\hom_{\mathcal{O}_X}(\mathcal{F}^{\prime\prime},\mathcal{G}))(U)\,\,(\sharp\sharp); \end{gather} in particular: \begin{equation} \forall x\in X,\,0\to(\hom_{\mathcal{O}_X}(\mathcal{F}^{\prime\prime},\mathcal{G}))_x\to(\hom_{\mathcal{O}_X}(\mathcal{F},\mathcal{G}))_x\to(\hom_{\mathcal{O}_X}(\mathcal{F}^{\prime\prime},\mathcal{G}))_x\,\,(\sharp\sharp\sharp) \end{equation} are left-exact sequence of $\mathcal{O}_{X,x}$-modules, by the same reasoning.

In other words: $(*)$ implies $(\sharp\sharp\sharp)$, that is the claim!