$\text{Hom}_{\mathcal{O}_X}(\cdot, \mathcal{G})$ 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
Commented Feb 9, 2016 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 $Z = W$. Commented Feb 9, 2016 at 18:50
• Here I gave an explanation. Commented Nov 25, 2022 at 9:20

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})$$, at stalks level one has: $$$$\forall x\in X,\,0\to\hom_{\mathcal{O}_{X,x}}(\mathcal{F}^{\prime\prime}_x,\mathcal{G}_x)\to\hom_{\mathcal{O}_{X,x}}(\mathcal{F}_x,\mathcal{G}_x)\to\hom_{\mathcal{O}_{X,x}}(\mathcal{F}^{\prime}_x,\mathcal{G}_x)\,\,(**)$$$$ because the (contravariant) functors $$\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}$$-modules; applying the (contravariant) right exact functors $$\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}$$ in particular: $$$$\forall x\in X,\,0\to(\hom_{\mathcal{O}_X}(\mathcal{F}^{\prime\prime},\mathcal{G}))_x\to(\hom_{\mathcal{O}_X}(\mathcal{F},\mathcal{G}))_x\to(\hom_{\mathcal{O}_X}(\mathcal{F}^{\prime\prime},\mathcal{G}))_x\,\,(\sharp\sharp\sharp)$$$$ are left-exact sequence of $$\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? Commented Nov 17, 2019 at 9:02
• Why is the first sequence in $(\sharp\sharp)$ exact? Isn't that precisely what we want to show? Commented Nov 25, 2022 at 9:18
• Because the functor $\hom_{\mathcal{O}_{X|U}}\left(\_,\mathcal{G}\right)$ is a contravariat right-exact functor, as I wrote! Commented Nov 25, 2022 at 14:34