# Clarification in definition of locally free sheaf

Let $(X,\mathcal{O}_X)$ be a ringed space. Hartshorne defines a sheaf of $\mathcal{O}_X$ - modules as a sheaf $\mathcal{F}$ on $X$ such that for each open set $U\subseteq X$, the group $\mathcal{F}(U)$ is an $\mathcal{O}_X(U)$ - module and for any open set $V \subseteq U$ the restriction homomorphism $\mathcal{F}(U) \rightarrow \mathcal{F}(V)$ is compatible with the module structure via the ring homomorphism $\mathcal{O}_X(U) \rightarrow \mathcal{O}_X(V)$.

A locally free sheaf on $X$ is a sheaf $\mathcal{F}$ of $\mathcal{O}_X$ - modules such that $X$ can be covered by open sets $U_i$ for which $\mathcal{F}|_{U_i}$ is a free $\mathcal{O}_X|_{U_i}$ - module ( isomorphic to a direct sum of copies of $\mathcal{O}_X|_{U_i}$)

My question is why do we need that $\mathcal{F}$ is a sheaf of modules in the definition of locally free sheaf? If we can define an $\mathcal{O}_X|_{U_i}$ module structure on each $\mathcal{F}|_{U_i}$ then can't we use the sheaf properties to patch a module structure on $\mathcal{F}(U)$ where $U\subseteq X$ is any open set.

More precisely, can I cover $U$ by $U_i \cap U$ and define the $\mathcal{O}_X$ - module structure on $\mathcal{F}(U)$ by patching the $\mathcal{O}_X(U_i \cap U)$ - module structures of $\mathcal{F}(U_i \cap U)$ via the sheaf properties of $\mathcal{F}$? Is there something wrong with this approach? If not then why do we need the condition that $\mathcal{F}$ is a sheaf of modules? Isn't it a consequence of locally free?

I am really confused with this. Thanks in advance!

• You're dealing with objects of some category $\mathcal{C}$ and you say that those which satisfy some condition are called "blahs". Your question is "can I use the fact that a thing is a blah to show that it's an object of category $\mathcal{C}$". Whether or not you can, you're kind of putting the cart before the horse here. – KReiser Oct 6 '15 at 8:07