The Stacks project

Proof. Let $(M', M_1, \alpha _1)$ be a glueing datum. We will show that $M = H^0((M', M_1, \alpha _1))$ is a glueable for $(R \to R', f)$ and that $(M', M_1, \alpha _1) \cong \text{Can}(M)$.

We first check that the map $\text{d} : M' \oplus M_1 \to (M')_ f$ used in the definition of the functor $H^0$ is surjective. Observe that $(x, y) \in M' \oplus M_1$ maps to $\text{d}(x, y) = x/1 - \alpha _1^{-1}(y \otimes 1)$ in $(M')_ f$. If $z \in (M')_ f$, then we can write $\alpha _1(z) = \sum y_ i \otimes g_ i$ with $g_ i \in R'$ and $y_ i \in M_1$. Write $\alpha _1^{-1}(y_ i \otimes 1) = y_ i'/f^ n$ for some $y'_ i \in M'$ and $n \geq 0$ (we can pick the same $n$ for all $i$). Write $g_ i = a_ i + f^ n b_ i$ with $a_ i \in R$ and $b_ i \in R'$. Then with $y = \sum a_ i y_ i \in M_1$ and $x = \sum b_ i y'_ i \in M'$ we have $\text{d}(x, -y) = z$ as desired.

Since $M = H^0((M', M_1, \alpha _1)) = \mathop{\mathrm{Ker}}(\text{d})$ we obtain an exact sequence of $R$-modules

We will prove that the maps $M \to M'$ and $M \to M_1$ induce isomorphisms $M \otimes _ R R' \to M'$ and $M \otimes _ R R_ f \to M_1$. This will imply that $M$ is glueable for $(R \to R', f)$ and $\text{Can}(M) \cong (M', M_1, \alpha _1)$ as desired.

Since $f$ is a nonzerodivisor on $M_1$, we have $M[f^\infty ] \cong M'[f^\infty ]$. This yields an exact sequence

Since $R \to R_ f$ is flat, we may tensor this exact sequence with $R_ f$ to deduce that $M \otimes _ R R_ f = (M/M[f^\infty ]) \otimes _ R R_ f \to M_1$ is an isomorphism.

By Lemma 15.90.15 we have $\text{Tor}_1^ R(R', \mathop{\mathrm{Coker}}(M' \to (M')_ f)) = 0$. The sequence (15.90.16.2) thus remains exact upon tensoring over $R$ with $R'$. Using $\alpha _1$ and Lemma 15.88.8 the resulting exact sequence can be written as

This yields an isomorphism $(M/M[f^\infty ]) \otimes _ R R' \cong M'/M'[f^\infty ]$. This implies that in the diagram

the third vertical arrow is an isomorphism. Since the rows are exact and the first vertical arrow is an isomorphism by Lemma 15.88.8 and $M[f^\infty ] = M'[f^\infty ]$, the five lemma implies that $M \otimes _ R R' \to M'$ is an isomorphism.

The above shows that $\text{Can}$ is essentially surjective and that the functor $H^0$ maps into the category of glueable modules. Due to the exactness of (15.90.9.1) for glueable modules we have $H^0 \circ \text{Can} = \text{id}$ on the category of glueable modules. This implies $\text{Can}$ is fully faithful by Lemma 15.89.11 combined with Categories, Lemma 4.24.4. This finishes the proof. $\square$