TITLE: When is the morphism from global section to stalk surjective
QUESTION [9 upvotes]: Let $X$ be a projective scheme over $\mathbb{C}$ and $\mathcal{F}$ a locally free sheaf on $X$ of rank $2$. Take a point $p$ in the support of $\mathcal{F}$. Suppose that there exists at least $3$ linearly independent global sections none of which vanish at this point $p$. Is it then true that the natural morphism from $H^0(\mathcal{F})$ to the stalk $H^0(\mathcal{F}_p)$ is surjective? If not, is there any bound on the number of linearly independent global sections as above which will ensure a positive answer?

REPLY [6 votes]: I do not think this is true. Since the question is local, we may assume $X = \operatorname{Spec}(A)$ is affine and $\mathcal F$ is free on $X$. Thus, $\mathcal F = \mathcal O_X\oplus\mathcal O_X$, the global sections restrict to $H^0(X, \mathcal F) = A\oplus A$, and $\mathcal F_p = A_p\oplus A_p$. The natural map factors through $A\oplus A\to A_p\oplus A_p$, which is certainly not always surjective.
For example, if we start with $X = \mathbb P^1$ and $\mathcal F = \mathcal O(1)\oplus\mathcal O(1)$, then the support is all of $\mathbb P^1$ and there are linearly independent global sections $(x,0),(y,0),(0,x),(0,y)$, where $x,y$ are coordinates. But in an affine chart with coordinate $u$, setting $p = 0$, we end up with the map $k[u]\oplus k[u]\to k[u]_{(u)}\oplus k[u]_{(u)}$ which is not surjective.
Note that by playing this game with higher twists on $\mathbb P^1$, we can get any number of linearly independent global sections, but the local picture will remain unchanged.