Lemma 15.89.11. In Remark 15.89.10 the functor $H^0 : \text{Glue}(R \to S, f_1, \ldots , f_ t) \to \text{Mod}_ R$ is a right adjoint to the functor $\text{Can} : \text{Mod}_ R \to \text{Glue}(R \to S, f_1, \ldots , f_ t)$.

Proof. Let $\mathbf{M} = (M', M_ i, \alpha _ i, \alpha _{ij})$ be an object of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$. For any $R$-module $N$ there is a map

$\mathop{\mathrm{Hom}}\nolimits _{\text{Glue}(R \to S, f_1, \ldots , f_ t)}(\text{Can}(N), \mathbf{M}) \to \mathop{\mathrm{Hom}}\nolimits _ R(N, H^0(\mathbf{M}))$

sending $\psi$ to $H^0(\psi )$ composed with the obvious map $N \to H^0(\text{Can}(N))$. By construction the displayed map is an isomorphism for $N = R$ (even if $R \to H^0(\text{Can}(R))$ is not an isomorphism in general). The category $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ has direct sums and cokernels. The functor $\text{Can}$ commutes with direct sums and cokernels. From these observations we find the displayed map is bijective by writing $N$ as a cokernel of a map between free $R$-modules. We omit the details. $\square$

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