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Let $C$ be a smooth projective irreducible curve. Let $Z$ be a closed subscheme of $C$ consisting of a finite set of points. Denote by $i:Z \to C$ the closed immersion. Note that this is a proper morphism. Let $\mathcal{F}$ be a locally free sheaf on $C$. So, $i_*i^* \mathcal{F}$ is a coherent sheaf. Hence, $H^0(i_*i^* \mathcal{F})$ is finite dimensional. If $\mathcal{F}$ is globally generated or ample, can we expect that the natural morphism $$H^0(\mathcal{F}) \to H^0(i_*i^* \mathcal{F})$$ is surjective?

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up vote 2 down vote accepted

The sources has some fixed dimension, whereas the target is equal to the direct sum, over $z \in Z$, of the fibre of $\mathcal F$ at $z$. So if the number of points in $Z$ is sufficienly large, the map can't be surjective.

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If the number of points is just $1$, can it be surjective? – user46578 Feb 8 '14 at 13:43
@user46578: yes, if $Z$ is just 1 point and $F$ is globally generated, then it will be surjective. – user64687 Feb 10 '14 at 11:30
@Dubh: Why? Could you give a reference/ idea of proof. You could write it as an answer. – user46578 Feb 10 '14 at 11:44
@user46578: Dear user, If $\mathcal F$ is globally generated, then its global sections span the fibre of $\mathcal F$ at the point of $Z$ (by definition), and so the restriction map $H^0(\mathcal F) \to H^0(i_* i^* \mathcal F)$ is surjective. Regards, – Matt E Feb 10 '14 at 12:12
@Matt E: Thanks – user46578 Feb 10 '14 at 12:24

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