About globally generated Sheaves On Vakil's Lecture Notes, he puts an important  exercise that says:
''Suppose $\cal{F}$ is a finite type quasicoherent sheaf on a scheme $X$. Show that$\cal{F}$ is globally generated at $p$ if and only if “the fiber of $\cal{F}$ is generated
by global sections at $p$”, i.e., the map from global sections to the fiber $\cal{F}_p/\cal{m}_p\cal{F}_p$ is surjective, where $\cal{m}_p$ is the maximal ideal of $\cal{O}_{X,p}$''. 
Though he said that this is an easy exercise I can not do this,  I'm a little confuse with the term ''the map from global sections to the fiber $\cal{F}_p/\cal{m}_p\cal{F}_p$ is surjective'' what this in fact means, that I have to find a surjective map of modules $\cal{F}(X) \to \cal{F}_p/\cal{m}_p\cal{F}_p$? 
If yes, he says that we have to use the Geometric Nakayma Lemma, but I don't see how to use this here. 
Has anyone have a clue?
Thank you so much!
 A: Allow me to use $x$ instead of $p$ (force of habit). I had to look up the definition in Vakil's notes of "generated by global sections at a point $x$" as I had not seen this before. This means that there is a map of sheaves of modules $\mathscr{O}_X^I\to\mathscr{F}$ (for some set $I$) which induces a surjection on stalks at $x$. This is not the same as what you write (..."the map from global sections..."). What this means is that there is a set of global sections $\{f_i:i\in I\}\subseteq\mathscr{F}(X)$ whose images in the stalk $\mathscr{F}_x$ span $\mathscr{F}_x$ as an $\mathscr{O}_{X,x}$-module.
Okay, now that we have the definition straight, there is an issue with this exercise. One cannot expect that the map $\mathscr{F}(X)\to\mathscr{F}_x/\mathfrak{m}_x\mathscr{F}_x$ is surjective. I think the right exercise is as follows: let $M\subseteq\mathscr{F}_x$ be the $\mathscr{O}_{X,x}$-submodule of $\mathscr{F}_x$ spanned by the image of $\mathscr{F}(X)\to\mathscr{F}_x$. Then $M=\mathscr{F}_x$  (this equality is literally equivalent to $\mathscr{F}$ being generated by global sections at $x$ in the sense defined above) if and only if the map $M/\mathfrak{m}_xM\to\mathscr{F}_x/\mathfrak{m}_x\mathscr{F}_x$ is surjective, or equivalently, $M+\mathfrak{m}_x\mathscr{F}_x=\mathscr{F}_x$. But now this is immediate. Indeed, because $\mathscr{F}$ is of finite type, $\mathscr{F}_x$ is a finitely generated $\mathscr{O}_{X,x}$-module, and Nakayama's lemma is applicable. It says that $M=\mathscr{F}_x$ if and only if $\mathfrak{m}_x(\mathscr{F}/M)=\mathscr{F}/M$, or equivalently, if and only if $M+\mathfrak{m}_x\mathscr{F}_x=\mathscr{F}_x$.
We don't actually need to assume $X$ is a scheme, just a locally ringed space, and we also don't need to assume that $\mathscr{F}$ is quasi-coherent. 
Here's an example showing the exercise as originally stated isn't right. Consider $X=\mathrm{Spec}(\mathbf{Z})$ and the finite type (quasi-coherent) $\mathscr{O}_X$-module associated to the $\mathbf{Z}$-module $\mathbf{Z}$ itself, i.e., just $\mathscr{O}_X$! This is certainly generated by global sections. However, look at the stalk at the generic point $x=(0)$. This is $\mathbf{Z}_{(0)}=\mathbf{Q}$, and its ``fiber" is also just $\mathbf{Q}$. The map $\mathbf{Z}=\mathscr{O}_X(X)\to\mathscr{O}_{X,x}/\mathfrak{m}_x\mathscr{O}_{X,x}=\mathbf{Q}$ is definitely not surjective, but of course, the image certainly spans $\mathbf{Q}$ as a module over $\mathscr{O}_{X,x}$, i.e., as a $\mathbf{Q}$-vector space. 
