A canonical homomorphism of sheaves of modules

Question

Let $\mathcal F$ be a sheaf of $\mathcal O_X$ modules on a scheme $X$. Fix an affine open subset $U$. If $M$ is a module over the coordinate ring of $U$, we let $\tilde{M}$ denote the associated sheaf of modules.

Why do we have a morphism of sheaves

$$\mathcal F (U)^\tilde{} \rightarrow \mathcal F|_U?$$

Of course on the set $U$ this is obvious (it's the identity), but I don't see how to define the morphism on subsets $V\subset U$. It suffices to do it just on principal open subsets. On these, we know that $ F (U)^\tilde{}$ has a nice description: on $D(f)$, it is $F (U)^\tilde{}_f$. But I don't know what $\mathcal F_U$ is on such sets, so I don't see how to define the homomorphism.

(Perhaps I am missing something obvious -- it is quite late...)

A reference to this morphism appears on page 160 of Liu.

Answer 1

Note that it suffices to take $X = \mathrm{Spec}(A)$ and to construct a homomorphism $$ (\Gamma(X, \mathscr{F}))^\sim \longrightarrow \mathscr{F}. $$ On an open subset $U = D(f)$ $(f \in A)$, the homomorphism $$ \Gamma(X, \mathscr{F})_f \to \Gamma(D(f), \mathscr{F}) $$ is induced by the restriction homomorphism $\Gamma(X,\mathscr{F}) \to \Gamma(U,\mathscr{F})$. To see this, note that $U$ is precisely the set of points $x \in X$ such that $f_x \in \mathscr{O}_x$ is invertible.

Answer 2

More generally, if $X$ is an arbitrary ringed space, $M$ is a module over $\mathcal{O}_X(X)$, then $\tilde{M} := M \otimes_{\mathcal{O}_X(X)} \mathcal{O}_X$, the sheaf associated to the presheaf $U \mapsto M \otimes_{\mathcal{O}_X(X)} \mathcal{O}_X(U)$, is a sheaf of modules on $X$, and it enjoys the universal property that $M \mapsto \tilde{M}$ is left adjoint to the section functor. See Stacks Project, Lemma 16.10.5. The unit of this adjunction is a natural map $\widetilde{F(X)} \to F$, where $F$ is a sheaf of modules on $X$. It is induced by the restriction maps $F(X) \otimes_{\mathcal{O}_X(X)} \mathcal{O}_X(U) \to F(U)$, $s \otimes \lambda \mapsto s|_U \lambda$.