# How to show that a sheaf is itself a sheaf of modules?

I am currently writing the proof for the following proposition: **Show that the sheaf of sections on a vector bundle $$V$$ over $$X$$ is a sheaf of modules over a sheaf of continuous function on $$X$$. **

I have so far:

Let $$\pi: V \to X$$ be the bundle projection. Let $$\mathscr{L}_X$$ be the sheaf of sections such that for $$U \subset X$$, $$\mathscr{L}_X (U)$$ is the set of all functions $$\sigma: U \to \pi^{-1}U$$. Let $$\mathscr{{T}}_X$$ be the sheaf of continuous functions on $$X$$ such that $$\mathscr{T}_X (U)$$ is the ring of all maps $$U \to V$$. Finally, let $$\mathscr{G}_X$$ be the sheaf which associates a $$\mathscr{T}_X (U)$$-module to every $$U$$. Now, it suffices to show that $$\mathscr{L}_X$$ is a sheaf of $$\mathscr{T}_X$$-modules.

I am confused as to how to show that a sheaf is itself a sheaf of modules.

Let $$X$$ be a space and $$\mathcal{O}_X$$ a sheaf of rings (thought of as continuous, smooth, holomorphic, or algebraic functions). A sheaf of Abelian groups $$\mathcal{F}$$ is a sheaf of $$\mathcal{O}_X$$-modules if there are multiplication maps $$\mathcal{O}_X(U)\times \mathcal{F}(U)\to \mathcal{F}(U)$$ for each open $$U\subset X$$ compatible with restrictions in the obvious way.
In this case, what you need to show is that given a section $$s$$ of $$V$$ over $$U\subset X$$ and a continuous function $$f$$ on $$U$$ (say here valued in $$\Bbb{R}$$ or $$\Bbb{C}$$) there is a way to define $$(f,s)\mapsto f\cdot s$$ in such a way that it is compatible with the restriction maps. In this case, simply define a new section of $$V$$ over $$U$$ by $$(f\cdot s)(p) = f(p)\cdot s(p).$$