Morphism of $\mathcal{O}_{X}$-modules on an affine scheme.

Question

Suppose that we have an affine scheme $X=\operatorname{Spec}(R)$, and suppose that for all distinguished opens $X_{f}(=D(f))$ for $f\in R$ we have morphisms of $R_{f}$-modules $$\varphi_{D(f)}:\mathcal{F}(D(f))\rightarrow\mathcal{G}(D(f)).$$ What do we have to check to conclude that these maps patch together to a morphism of $\mathcal{O}_{X}$-modules $$\varphi:\mathcal{F}\rightarrow\mathcal{G}?$$

Answer 1

All you need is compatibility with restriction. Remember that the sections of a sheaf (with values in a sufficiently nice category) are determined by their sections on the open subsets of a basis for the topology: if $U$ is an open set with a covering by $\{U_i\}_{i\in I}$, then $\mathcal{F}(U)$ is the equalizer of the diagram $$\prod_{i\in I} \mathcal{F}(U_i)\rightrightarrows\prod_{i,j\in I}\mathcal{F}(U_i\cap U_j).$$ If the $U_i$ are elements of a base, then $U_i\cap U_j$ are too, and thus if we have compatible restriction morphisms $\mathcal{F}(U_i)\to\mathcal{G}(U_i)$ for all elements of a basis for the topology, we get an induced morphism on the equalizers of the above diagram for any $U$, and these maps $\mathcal{F}(U)\to\mathcal{G}(U)$ are compatible by the assumption that the maps on the basis sets were. So since the distinguished affine opens $D(f)$ form a basis for the topology on an affine scheme, we get a morphism of sheaves assuming our maps are compatible.