# Why is sheaf Hom left-exact?

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.

• 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. – Mariano Suárez-Álvarez Feb 1 '17 at 6:28
• The lemma above your cited lemma in the Stacks Project says that sheaf Hom is right adjoint to the tensor product. – MooS Feb 1 '17 at 6:31
• But the sheaf Hom is not a Hom-set in an abelian category. – Monstrous Moonshine Feb 1 '17 at 6:32
• But right adjoints are always left exact... – MooS Feb 1 '17 at 6:37
• @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. – Devlin Mallory Sep 4 '17 at 16:39