The Stacks project

Proof. Proof of (1). Say $I = (f_1, \ldots , f_ t)$. It is clear that the localization $\varphi _{f_ i}$ is an isomorphism. Thus we see that $(M', N_{f_ i}, \varphi _{f_ i}, can_{ij})$ is an object of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$, see Remark 15.89.10. By Proposition 15.89.16 we conclude that there exists an $R$-module $M$ such that $M' = M \otimes _ R S$ and $N_{f_ i} = M_{f_ i}$ compatibly with the isomorphisms $\varphi _{f_ i}$ and $can_{ij}$. There is a morphism

of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ which uses $\varphi $ in the first component. This corresponds to an $R$-module map $\psi : M \to N$ (by the equivalence of categories of Proposition 15.89.16). The composition of the base change of $M \to N$ with the isomorphism $M' \cong M \otimes _ R S$ is $\varphi $, in other words $M \to N$ is compatible with $\varphi $.

Proof of (2). This is just the dual of the argument above. Namely, the localization $\varphi _{f_ i}$ is an isomorphism. Thus we see that $(N', M_{f_ i}, \varphi _{f_ i}^{-1}, can_{ij})$ is an object of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$, see Remark 15.89.10. By Proposition 15.89.16 we conclude that there exists an $R$-module $N$ such that $N' = N \otimes _ R S$ and $N_{f_ i} = M_{f_ i}$ compatibly with the isomorphisms $\varphi _{f_ i}^{-1}$ and $can_{ij}$. There is a morphism

of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ which uses $\varphi $ in the first component. This corresponds to an $R$-module map $\psi : M \to N$ (by the equivalence of categories of Proposition 15.89.16). The composition of the base change of $M \to N$ with the isomorphism $N' \cong N \otimes _ R S$ is $\varphi $, in other words $M \to N$ is compatible with $\varphi $.

The final statement follows for example from Lemma 15.89.3. $\square$