Sheaf associated to a locally free presheaf of modules

Question

Just checking here. It is true isn't it, that the sheaf associated to a locally free presheaf of $O_X$ modules (I suppose there is an example that is not a sheaf??) over a scheme $X$ is a locally free sheaf of $O_X$ modules? If the associated sheaf is coherent then this just textbook (e.g. Hartshorne Exercise II.5.7b), but otherwise I still would suppose this is true simply by definition of locally free presheaf (i.e. free when restricted to an open set)

Answer 1

Suppose $\mathcal{F}$ is a presheaf of $\mathcal{O}_X-$modules that is locally free. This means that there exists an open cover $\{U_\alpha\}$ such that there exists $\varphi_\alpha:\mathcal{F}|_{U_\alpha}\xrightarrow{\sim}\bigoplus_{I}\mathcal{O}_{U_\alpha}$ for each $\alpha$. Such a map is an isomorphism on stalks. We know that $\mathcal{F}|_{U_\alpha}$ can be regarded as a presheaf on $U_\alpha$, and then $\mathcal{F}|_{U_\alpha}^+\cong \mathcal{F}^+|_{U_\alpha}$. By the universal property of the sheafification, the isomorphism $\varphi_\alpha:\mathcal{F}|_{U_\alpha}\to \bigoplus_I \mathcal{O}_{U_\alpha}$ induces a map $\mathcal{F}^+|_{U_\alpha}\to \bigoplus_I \mathcal{O}_{U_\alpha}.$ This map is an isomorphism: it can be checked on stalks but we know that the stalk maps were isomorphisms to begin with since $\varphi_\alpha$ was an isomorphism. So, $\mathcal{F}^+|_{U_\alpha}\cong \bigoplus_I\mathcal{O}_{U_\alpha}$.

Actually, this wasn't strictly necessary. It suffices to note that since $\mathcal{F}|_{U_\alpha}$ is isomorphic to a sheaf, it is itself a sheaf on $U_\alpha$. Hence, $\mathcal{F}^+|_{U_\alpha}\cong \mathcal{F}|_{U_\alpha}\cong \bigoplus_I \mathcal{O}_{U_\alpha}.$

By the way, the tensor product of sheaves needs to be sheafified to get a sheaf in general, so I think you can construct an example of a locally free presheaf that is not a sheaf by tensoring a pair of locally free sheaves.