I want to prove for an $\mathcal{R}-$module $\mathcal{F}$ over a topological space $X$,where $R$ is a sheaf of rings,if there exist sections $s_1,\ldots,s_p\in\mathcal{F}(X)$ that generate every stalk $\mathcal{F}_x$ over $\mathcal{R}_x$,then there exists an exact sequence $\mathcal{R}^p\xrightarrow{\varphi} \mathcal{F}\rightarrow 0$ of $\mathcal{R}-$module.
Here is my idea.$\forall f\in\mathcal{F}(X),\exists s_1,\ldots,s_p$ such that $\forall x\in X,\exists a_{1x},\ldots,a_{px}$ such that $f_x=a_{1x}s_{1x}+\cdots+a_{px}s_{px}$.In another words,$\exists U\in\mathcal{U}_x$ such that $f|U=a_{1U}s_{1U}+\cdots+a_{pU}s_{pU}$.Then we obtain a collection of local data $(\{U_\alpha\},a_{iU_\alpha})$.But how to glue these datum to get a global section?Can you help me?Thank you.
I assume that $(X,R)$ is a ringed space. Recall that epimorphisms of $R$-modules can be detected locally, and that it is not true that epimorphisms are sectionwise epimorphisms. (Sheaf cohomology measures the obstruction for this.) That is, $ F \to G$ is an epimorphism iff for each local section $s \in G(U)$ there is an open covering $U = \cup_i U_i$ such that $s|_{U_i}$ has a preimage in $F(U_i)$ (but in general we cannot find a preimage of $s$ in $F(U)$).