We know, that $\mathbf{Hom}(\mathcal{F},\mathcal{G})$ is abelian group of all morphisms between $\mathcal{O}_X$-modules $\mathcal{F}$ and $\mathcal{G}$, and $\mathcal{Hom}(\mathcal{F},\mathcal{G})$ is sheaf, which section $\mathcal{Hom}(\mathcal{F},\mathcal{G})(U)$ is $\mathcal{O}_X(U)$-module of all morphisms between module $\mathcal{F}(U)$ and module $\mathcal{G}(U)$.

So, for me statement $\Gamma(\mathcal{Hom}(\mathcal{F},\mathcal{G}))=\mathbf{Hom}(\mathcal{F},\mathcal{G})$ sounds like "every morphism between two $\mathcal{O}_X$-modules uniquely determined by morphism between global sections of them". It can't be true, right? So, where is mistake?


$\newcommand{\c}{\mathcal}$Your definition of $\c{Hom}(\c F,\c G)(U)$ is wrong. It is not given the $\c O_X(U)$-module morphisms $\c F(U)\to\c G(U)$, it is given by sheaf homomorphisms (actually $\c O_X|_U$-module morphisms) $\c F|_U\to\c G|_U$.

Your mistake is a normal one to make; at first glance it seems very natural to define $\c{Hom}(\c F,\c G)(U)=\hom_{\c O_X(U)}(\c F(U),\c G(U))$, but actually you'll find that if you take this definition, there is no natural way to define restriction morphisms, so $\c{Hom}(\c F,\c G)$ couldn't be a presheaf, let alone a sheaf.

Edit: also, I believe that your confusion seems to be coming from an awkward clash of notation. In the link you gave, where you say your question comes from, they are using $\c F(U)$ to denote the restriction sheaf of $\c F$ to $U$, rather than the sections of $\c F$ over $U$. This is not too uncommon, for instance this is the notation used in Serre's FAC (though that's one of the only places I've seen it so far).

  • $\begingroup$ Thank you, it totally clear now! $\endgroup$ – kp9r4d Mar 28 '17 at 19:59
  • $\begingroup$ @kp9r4d glad I could help! $\endgroup$ – Alex Mathers Mar 28 '17 at 20:02

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