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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.

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1 Answer 1

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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 $(f\cdot s)(p) = f(p)\cdot s(p).$

I leave it to you to write out the details.

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