For a line bundle $L$ on a curve $C$ the base locus of $L$, or equivalently the locus where the evaluation map $$H^0(L)\otimes \mathcal{O}_C\to L$$ fails to be surjective is a proper closed subset of $C$. That is, the above map is always (at least) generically surjective. I am wondering if that holds for a vector bundle as well. Precisely, if $E$ is a vector bundle of rank $r$ such that $h^0(E)\geq r$, then is it true that the map $$H^0(E)\otimes \mathcal{O}_C\to E$$ is generically surjective?
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2$\begingroup$ The invertible sheaf $L$ is generically globally generated only if $h^0(C,L)$ is at least $1$. There are many examples of higher rank locally free sheaf $E$ where $h^0(C,E)$ is at least $r$, yet the sheaf is not generically globally generated. For instance, for $C$ equal to $\mathbb{P}^1$, for $E$ equal to $\mathcal{O}(-a)\oplus \mathcal{O}(b)$ with $a,b>0$, this happens. $\endgroup$– Jason StarrCommented Jul 17, 2017 at 14:44
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First, I assume that you meant $h^0(L)\geq 1$ in the first statement, else it is clearly false. Second one is false too, since you can take $E$ to be the direct sum of a line bundle $L$ with $h^0(L\geq 2$ and another one $M$, with $h^0(M)=0$. Then the above map is not generically surjective.
