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

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.

• If $\delta_W$ is surjective it is right cancellable. – Pedro Tamaroff Feb 9 '16 at 2:23
• @Pedro The problem is that $\delta_W$ is only cancellable under the limit; meaning: for any $b \in \mathcal{F''}(W)$, you can find some $a \in \mathcal{F}(Z)$ for some open sets $Z \ni x$ such that the image of $\delta_Z(s) \in \mathcal{F'}(Z)$ and $b \in \mathcal{F''}(W)$ becomes equal in some $T \subseteq Z, W$ under the restriction homomorphism of $\mathcal{F''}$. There is no guarantee that $Z = W$. – An Hoa Feb 9 '16 at 18:50

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 $$\forall x\in X,\,\mathcal{F}^{\prime}_x\to\mathcal{F}_x\to\mathcal{F}^{\prime\prime}_x\to0\,\,(*)$$ are right-exact sequences of $\mathcal{O}_{X,x}$-modules.
Applying $\hom_{\mathcal{O}_X}(\_,\mathcal{G})$, at stalks level one has: $$\forall x\in X,\,0\to\hom_{\mathcal{O}_{X,x}}(\mathcal{F}^{\prime\prime}_x,\mathcal{G}_x)\to\hom_{\mathcal{O}_{X,x}}(\mathcal{F}_x,\mathcal{G}_x)\to\hom_{\mathcal{O}_{X,x}}(\mathcal{F}^{\prime}_x,\mathcal{G}_x)\,\,(**)$$ because the (controvariant) functors $\hom_{\mathcal{O}_{X,x}}(\_,\mathcal{G}_x)$ are right-exact, the sequences $(**)$ are left-exact.
By $(*)$ one can affirm that $$\forall U\subseteq X\,\text{open},\,\mathcal{F}^{\prime}_{|U}\to\mathcal{F}_{|U}\to\mathcal{F}^{\prime\prime}_{|U}\to0\,\,(\sharp)$$ are right-exact sequences of $\mathcal{O}_{X|U}$-modules; applying the (controvariant) 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: $$\forall x\in X,\,0\to(\hom_{\mathcal{O}_X}(\mathcal{F}^{\prime\prime},\mathcal{G}))_x\to(\hom_{\mathcal{O}_X}(\mathcal{F},\mathcal{G}))_x\to(\hom_{\mathcal{O}_X}(\mathcal{F}^{\prime\prime},\mathcal{G}))_x\,\,(\sharp\sharp\sharp)$$ 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!
• Don't worry! If I'm not wrong: $(**)$ is redundant. Since it is another result, I didn't delete it. Am I clear? – Armando j18eos Nov 17 '19 at 9:02