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Fix a ringed space $(X,\mathcal{O}_X)$ and two sheaves of $\mathcal{O}_X$-modules $\mathscr{F}$ and $\mathscr{G}$, then we can define the sheaf $\mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{F},\mathscr{G})$ by setting $$\mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{F},\mathscr{G})(U)=\mathrm{Hom}_{\mathcal{O}_X\vert_U}(\mathscr{F}\vert_U,\mathscr{G}\vert_U)$$ with the obvious restriction maps, so that using maps of $\mathcal{O}_X$-modules $\varphi:\mathscr{G}\to\mathscr{G}'$ and $\psi:\mathscr{F}'\to\mathscr{F}$ we can induce maps $$\mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{F},\mathscr{G})\to \mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{F},\mathscr{G}')$$ and $$\mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{F},\mathscr{G})\to\mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{F'},\mathscr{G})$$ given on the open set $U$ by $$f\mapsto \varphi\vert_U\circ f$$ and $$f\mapsto f\circ\psi\vert_U$$ respectively. Now, Lemma 17.20.12 of the Stacks Project claims that any short exact sequence $$0\to\mathscr{F}_1\to\mathscr{F}_2\to\mathscr{F}_3\to 0$$ induces exact sequences $$0\to\mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{F_3},\mathscr{G})\to\mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{F_2},\mathscr{G})\to\mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{F_1},\mathscr{G})$$ and $$0\to\mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{G},\mathscr{F_1})\to\mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{G},\mathscr{F_2})\to\mathscr{H}\!\mathit{om}_{\mathcal{O}_X}(\mathscr{G},\mathscr{F_3})$$ Now, I can prove that the first (nontrivial) map is injective in both cases, but I don't see how to show exactness in the middle.

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    All Homs in all abelian categories are left exact. This has nothing to do with ringed spaces or sheaves. I suggest you look at this at the level of generality of abeliancategories categories, to remove all irrelevant details. – 2017-02-01
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    The lemma above your cited lemma in the Stacks Project says that sheaf Hom is right adjoint to the tensor product. – 2017-02-01
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    But the sheaf Hom is not a Hom-set in an abelian category. – 2017-02-01
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    But right adjoints are always left exact... – 2017-02-01
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    Ok I understand now. – 2017-02-01
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    On a similar note, is the sheaf Ext balanced? – 2017-02-01
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    @Monstrous Moonshine What do you mean by balancing of Ext in this situation? Usually balancing of Ext means that $R^n \operatorname{Hom} (M,-) (N) \cong R^n \operatorname{Hom} (-,N) (M)$, when one works in a category with enough injectives and projectives. – 2017-02-01
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    @Monstrous Moonshine in this case, you can't obtain $\operatorname{Ext}^n (\mathcal{F},\mathcal{G})$ / $\mathcal{Ext}^n (\mathcal{F},\mathcal{G})$ by deriving $\operatorname{Hom} (-,\mathcal{G})$ / $\mathcal{Hom} (-,\mathcal{G})$, because we don't have enough projectives. – 2017-02-01
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    Still, there remains a question: If $\mathcal F$ admits a free or projective resolution (which can happen even if not all objects admit such), can you use it to compute $\mathcal{Ext}^n(\mathcal F,\mathcal G)$? That is a legit question. It is definitely wrong for $\operatorname{Ext}$ because we have $\operatorname{Hom}(\mathcal O_X,-)=\Gamma(X,-)$, which has non-trivial higher cohomology. But is it true for $\mathcal{Ext}$? I would say the answer is yes for free resolutions. – 2017-02-02
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    @MooS, a little late to this, but Hartshorne Prop. III.6.5 answers your last question: you can calculate the ext sheaves via a locally free free resolution. – 2017-09-04

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