In this Stacks Project entry: Section 17.9: Modules of finite type, the authors define a sheaf of $\mathcal{O}_X$-modules $\mathcal{F}$ (in which $(X,\mathcal{O}_X)$ is a ringed space) to be of finite type if for every $x\in X$, there is an open neighborhood $U$ such that $\mathcal{F}_{|U}$ is generated by finitely many sections. In other words, there is a surjective morphism of $\mathcal{O}_X$-modules
$$\bigoplus_{i=1,\dots,n}\mathcal{O}_U\longrightarrow \mathcal{F}_{|U}$$
In the language of Definition 17.4.1, we say that $\mathcal{F}_{|U}$ is generated by finitely many global sections.
I'm trying to understand Lemma 17.9.3 and Lemma 17.9.4, and I think Lemma 17.9.4 can be used to prove Lemma 17.9.3. Here is the statement of Lemma
17.9.4:
Let $X$ be a ringed space. Let $\varphi:\mathcal{G}\rightarrow \mathcal{F}$ be a homomorphism of $\mathcal{O}_X$-modules. Let $x\in X$. Assume $\mathcal{F}$ of finite type and the map on stalks $\varphi_x:\mathcal{G}_x\rightarrow \mathcal{F}_x$ surjective. Then there exists an open neighborhood $x\in U\subset X$ such that $\varphi_{|U}$ is surjective.
I'm struggling with the proof of this lemma. They take an open neighborhood $U$ of $x$ such that $\mathcal{F}$ is generated by finitely many sections, namely $s_1,\dots,s_n\in \mathcal{F}(U)$. Then they shrink $U$ so that there is $t_1,\dots, t_n\in \mathcal{G}(U), \varphi(t_i)=s_i$. This is possible, since we can take $t_{i,x}\in \mathcal{G}_x$ that $\varphi_x(t_{i,x})=s_{i,x}$ and take the intersection of open neighborhood $U_i$ on which the $t_i$ are defined. But in the next sentence, they say that this complete the proof of the lemma, and for me this is not very convincing.
If I was right, saying $\mathcal{F}_{|U}$ can be generated by finitely many sections doesn't mean that there are finitely many sections over $U$ that generate $\mathcal{F}(U)$ as a module. In fact, since colimit of sheaves is sheafification of colimit of presheaves, a section $s\in \mathcal{F}(U)$
is locally look like a linear combination of $s_1,\dots,s_n$, i.e. there exists an open covering $U=\bigcup U_j$ that $s_{|U_j}=\sum r_{ij} s_{i|U_j}$. Hence the fact that the $s_i$ are containing in the image of $\varphi_{|U}$ doesn't imply $\varphi_{|U}$ is surjective, even if we try to restrict to some $U_j$ (which are different for different sections $s$).
Is there another way to think about a sheaf of $\mathcal{O}_X$-modules of finite type make this kind of "local-global" problems easier to deal with? Am I just confused at some points?
Any answers to this question will help me a lot. Thanks in advance.
Best Answer
Note that saying $\varphi|_U$ is surjective does not mean that every section of $\mathcal{F}$ over (a subset of) $U$ is the image of some section of $\mathcal{G}$ under $\varphi$. It just means that the image sheaf $\varphi|_U(\mathcal{G}|_U)$ is all of $\mathcal{F}|_U$, and the image sheaf is defined by sheafifying the image presheaf. In other words, it just means that sections of $\mathcal{F}$ over (subsets of) $U$ are locally images of sections of $\mathcal{G}$ under $\varphi$, not globally.
Another way to think about it is that if $\mathcal{F}|_U$ is generated by global sections $s_1,\dots,s_n$, then any $\mathcal{O}_U$-submodule that contains all of $s_1,\dots,s_n$ must be all of $\mathcal{F}|_U$. Since the image sheaf $\varphi|_U(\mathcal{G}|_U)$ is such a submodule, it must be all of $\mathcal{F}|_U$. From this perspective, you don't have to worry about what all this means on the level of sections, as long as you only ever talk about $\mathcal{O}_U$-submodules.