Let $X$ be a quasi-projective irreducible scheme, $\mathcal{F}_1$ a globally generated $\mathcal{O}_X$-module and $\mathcal{F}_2$ a coherent sheaf over $X$. Suppose that $\mathcal{F}_1$ is globally generated by the sections $s_1,...,s_n \in \Gamma(X,\mathcal{F}_1)$. I would imagine that it follows from the definition of morphism of sheaves that any morphism from $\mathcal{F}_1$ to $\mathcal{F}_2$ is determined by the image of $s_1,...,s_n$ and these sections are mapped to elements in $\Gamma(X,\mathcal{F}_2)$. Is this true? More precisely, is it true that there exists an injective morphism from $\mbox{Hom}_X(\mathcal{F}_1,\mathcal{F}_2)$ to $\mbox{Hom}_{\mathcal{O}_X(X)}(\mathcal{F}_1(X),\mathcal{F}_2(X))$?
1 Answer
It is not true. Take for example $F_1$ and $F_2$ to be the structure sheaves of two distinct points on $X$. Then both are globally generated, the spaces of global sections are 1-dimensional, and there is a nontrivial morphism between these vector spaces. But there is no morphism between the sheaves.
On the other hand, the morphism $Hom(F_1,F_2) \to Hom(\Gamma(F_1),\Gamma(F_2)$ (the action of the functor of global sections $\Gamma$) is injective if $F_1$ is globally generated. Indeed, the condition ensures that the map $\Gamma(F_1)\otimes O_X \to F_1$ is surjective, hence applying $Hom(-,F_2)$ gives an injective morphism $$ Hom(F_1,F_2) \to Hom(\Gamma(F_1)\otimes O_X,F_2) = Hom(\Gamma(F_1),\Gamma(F_2)). $$
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$\begingroup$ Thanks for the answer. Can we at least get an injective morphism from $\mbox{Hom}_X(\mathcal{F}_1,\mathcal{F}_2)$ to $\mbox{Hom}_{\mathcal{O}_X(X)}(\mathcal{F}_1(X),\mathcal{F}_2(X))$? $\endgroup$ Jan 30, 2014 at 17:27
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$\begingroup$ Yes, injectivity holds, see the explanation. $\endgroup$– SashaJan 30, 2014 at 18:04