hcju/stacksqa huggingface dataset:
By Lemma \ref{lemma-lci-NL} the complexes $\NL_{B/A}$ and $\NL_{B'/A'}$ are perfect of tor-amplitude in $[-1, 0]$. Combining Lemmas \ref{lemma-tensor-NL}, \ref{lemma-pull-perfect}, and \ref{lemma-pull-tor-amplitude} we have $\NL_{B/A} \otimes_B B' = \NL_{B/A} \otimes_B^\mathbf{L} B'$ and this complex is also perfect of tor-amplitude in $[-1, 0]$. Choose a distinguished triangle $$ C \to \NL_{B/A} \otimes_B B'  \to \NL_{B'/A'} \to C[1] $$ in $D(B')$. By Lemmas \ref{lemma-two-out-of-three-perfect} and \ref{lemma-cone-tor-amplitude} we conclude that $C$ is perfect with tor-amplitude in $[-1, 1]$. By Lemma \ref{lemma-base-change-NL} the complex $C$ has only one nonzero cohomology module, namely the module of the lemma sitting in degree $-1$. This module is of finite presentation (Lemma \ref{lemma-n-pseudo-module}) and flat (Lemma \ref{lemma-tor-dimension}). Hence it is finite projective by Algebra, Lemma \ref{algebra-lemma-finite-projective}.
Let $(A_n)$ be an arbitrary inverse system. Let $(B_n)$ be the inverse system with $$ B_n = A_n \oplus A_{n - 1} \oplus \ldots \oplus A_1 $$ and transition maps given by projections. Let $A_n \to B_n$ be given by $(1, f_n, f_{n - 1} \circ f_n, \ldots, f_2 \circ \ldots \circ f_n$ where $f_i : A_i \to A_{i - 1}$ are the transition maps. In this way we see that every inverse system is a subobject of a ML system (Homology, Section \ref{homology-section-inverse-systems}). It follows from Derived Categories, Lemma \ref{derived-lemma-subcategory-right-acyclics} using Homology, Lemma \ref{homology-lemma-Mittag-Leffler} that every ML system is right acyclic for $\lim$, i.e., (3) holds. This already implies that $RF$ is defined on $D^+(\textit{Ab}(\mathbf{N}))$, see Derived Categories, Proposition \ref{derived-proposition-enough-acyclics}. Set $C_n = A_{n - 1} \oplus \ldots \oplus A_1$ for $n > 1$ and $C_1 = 0$ with transition maps given by projections as well. Then there is a short exact sequence of inverse systems $0 \to (A_n) \to (B_n) \to (C_n) \to 0$ where $B_n \to C_n$ is given by $(x_i) \mapsto (x_i - f_{i + 1}(x_{i + 1}))$. Since $(C_n)$ is ML as well, we conclude that (2) holds (by proposition reference above) which also implies (1). Finally, this implies by Derived Categories, Lemma \ref{derived-lemma-unbounded-right-derived} that $R\lim$ is in fact defined on all of $D(\textit{Ab}(\mathbf{N}))$. In fact, the proof of Derived Categories, Lemma \ref{derived-lemma-unbounded-right-derived} proceeds by proving assertions (4) and (5).
Let $K \in D(\textit{Ab}(\mathbf{N}))$ be the object represented by the system of complexes whose $n$th constituent is the complex $A^{-2}_n \to A^{-1}_n \to A^0_n \to A^1_n$. We will compute $H^0(R\lim K)$ using both spectral sequences\footnote{To use these spectral sequences we have to show that $\textit{Ab}(\mathbf{N})$ has enough injectives. A inverse system $(I_n)$ of abelian groups is injective if and only if each $I_n$ is an injective abelian group and the transition maps are split surjections. Every system embeds in one of these. Details omitted.} of Derived Categories, Lemma \ref{derived-lemma-two-ss-complex-functor}. The first has $E_1$-page $$ \begin{matrix} 0 & 0 & R^1\lim A^0_n  & R^1\lim A^1_n \\ A^{-2} & A^{-1} & A^0  & A^1 \end{matrix} $$ with horizontal differentials and all higher differentials are zero. The second has $E_2$ page $$ \begin{matrix} R^1\lim H^{-2}_n & 0 & R^1\lim H^0_n & R^1 \lim H^1_n \\ \lim H^{-2}_n & \lim H^{-1}_n & \lim H^0_n & \lim H^1_n \end{matrix} $$ and degenerates at this point. The result follows.
Suppose that for each $p$ the inverse system $(K_n^p)$ is right acyclic for $\lim$. By Lemma \ref{lemma-compute-Rlim} this gives a short exact sequence $$ 0 \to \lim_n K^p_n \to \prod\nolimits_n K^p_n \to \prod\nolimits_n K^p_n \to 0 $$ for each $p$. Since the complex consisting of $\lim_n K^p_n$ computes $R\lim K$ by Lemma \ref{lemma-compute-Rlim} we see that the lemma holds in this case. \medskip\noindent Next, assume $K = (K_n^\bullet)$ is general. By Lemma \ref{lemma-compute-Rlim} there is a quasi-isomorphism $K \to L$ in $D(\textit{Ab}(\mathbf{N}))$ such that $(L_n^p)$ is acyclic for each $p$. Then $\prod K_n^\bullet$ is quasi-isomorphic to $\prod L_n^\bullet$ as products are exact in $\textit{Ab}$, whence the result for $L$ (proved above) implies the result for $K$.
The long exact sequence of the distinguished triangle is $$ \ldots \to H^p(R\lim K) \to \prod\nolimits_n H^p(K_n^\bullet) \to \prod\nolimits_n H^p(K_n^\bullet) \to H^{p + 1}(R\lim K) \to \ldots $$ The map in the middle has kernel $\lim_n H^p(K_n^\bullet)$ by its explicit description given in the lemma. The cokernel of this map is $R^1\lim_n H^p(K_n^\bullet)$ by Lemma \ref{lemma-compute-Rlim}.
The assumption in particular implies that the pro-objects $H^p(E_n)$ and $H^p(D_n)$ are isomorphic. By the short exact sequences of Lemma \ref{lemma-break-long-exact-sequence} it suffices to show that given a map $(A_n) \to (B_n)$ of inverse systems of abelian groupsc which induces an isomorphism of pro-objects, then $\lim A_n \cong \lim B_n$ and $R^1\lim A_n \cong R^1\lim B_n$. \medskip\noindent The assumption implies there are $1 \leq m_1 < m_2 < m_3 < \ldots$ and maps $\varphi_n : B_{m_n} \to A_n$ such that $(\varphi_n) : (B_{m_n}) \to (A_n)$ is a map of systems which is inverse to the given map $\psi = (\psi_n) : (A_n) \to (B_n)$ as a morphism of pro-objects. What this means is that (after possibly replacing $m_n$ by larger integers) we may assume that the compositions $A_{m_n} \to B_{m_n} \to A_n$ and $B_{m_n} \to A_n \to B_n$ are equal to the transition maps of the inverse systems. Now, if $(b_n) \in \lim B_n$ we can set $a_n = \varphi_{m_n}(b_{m_n})$. This defines an inverse $\lim B_n \to \lim A_n$ (computation omitted). Let us use the cokernel of the map $$ \prod B_n \longrightarrow \prod B_n $$ as an avatar of $R^1\lim B_n$ (Lemma \ref{lemma-compute-Rlim}). Any element in this cokernel can be represented by an element $(b_i)$ with $b_i = 0$ if $i \not = m_n$ for some $n$ (computation omitted). We can define a map $R^1\lim B_n \to R^1\lim A_n$ by mapping the class of such a special element $(b_n)$ to the class of $(\varphi_n(b_{m_n}))$. We omit the verification this map is inverse to the map $R^1\lim A_n \to R^1\lim B_n$.
Set $B = \bigoplus_{i \in \mathbf{N}} (A_n)$ and hence $B = (B_n)$ with $B_n = \bigoplus_{i \in \mathbf{N}} A_n$. If $(A_n)$ is ML, then $B$ is ML and hence $R^1\lim A_n = 0$ and $R^1\lim B_n = 0$ by Lemma \ref{lemma-compute-Rlim}. \medskip\noindent Conversely, assume $(A_n)$ is not ML. Then we can pick an $m$ and a sequence of integers $m < m_1 < m_2 < \ldots$ and elements $x_i \in A_{m_i}$ whose image $y_i \in A_m$ is not in the image of $A_{m_i + 1} \to A_m$. We will use the elements $x_i$ and $y_i$ to show that $R^1\lim B_n \not = 0$ in two ways. This will finish the proof of the lemma. \medskip\noindent First proof. Set $C = (C_n)$ with $C_n = \prod_{i \in \mathbf{N}} A_n$. There is a canonical injective map $B_n \to C_n$ with cokernel $Q_n$. Set $Q = (Q_n)$. We may and do think of elements $q_n$ of $Q_n$ as sequences of elements $q_n = (q_{n, 1}, q_{n, 2}, \ldots)$ with $q_{n, i} \in A_n$ modulo sequences whose tail is zero (in other words, we identify sequences which differ in finitely many places). We have a short exact sequence of inverse systems $$ 0 \to (B_n) \to (C_n) \to (Q_n) \to 0 $$ Consider the element $q_n \in Q_n$ given by $$ q_{n, i} = \left\{ \begin{matrix} \text{image of }x_i &\text{if}& m_i \geq n \\ 0 & \text{else} \end{matrix} \right. $$ Then it is clear that $q_{n + 1}$ maps to $q_n$. Hence we obtain $q = (q_n) \in \lim Q_n$. On the other hand, we claim that $q$ is not in the image of $\lim C_n \to \lim Q_n$. Namely, say that $c = (c_n)$ maps to $q$. Then we can write $c_n = (c_{n, i})$ and since $c_{n', i} \mapsto c_{n, i}$ for $n' \geq n$, we see that $c_{n, i} \in \Im(C_{n'} \to C_n)$ for all $n, i, n' \geq n$. In particular, the image of $c_{m, i}$ in $A_m$ is in $\Im(A_{m_i + 1} \to A_m)$ whence cannot be equal to $y_i$. Thus $c_m$ and $q_m = (y_1, y_2, y_3, \ldots)$ differ in infinitely many spots, which is a contradiction. Considering the long exact cohomology sequence $$ 0 \to \lim B_n \to \lim C_n \to \lim Q_n \to R^1\lim B_n $$ we conclude that the last group is nonzero as desired. \medskip\noindent Second proof. For $n' \geq n$ we denote $A_{n, n'} = \Im(A_{n'} \to A_n)$. Then we have $y_i \in A_m$, $y_i \not \in A_{m, m_i + 1}$. Let $\xi = (\xi_n) \in \prod B_n$ be the element with $\xi_n = 0$ unless $n = m_i$ and $\xi_{m_i} = (0, \ldots, 0, x_i, 0, \ldots)$ with $x_i$ placed in the $i$th summand. We claim that $\xi$ is not in the image of the map $\prod B_n \to \prod B_n$ of Lemma \ref{lemma-compute-Rlim}. This shows that $R^1\lim B_n$ is nonzero and finishes the proof. Namely, suppose that $\xi$ is the image of $\eta = (z_1, z_2, \ldots)$ with $z_n = \sum z_{n, i} \in \bigoplus_i A_n$. Observe that $x_i = z_{m_i, i} \bmod A_{m_i, m_i + 1}$. Then $z_{m_i - 1, i}$ is the image of $z_{m_i, i}$ under $A_{m_i} \to A_{m_i - 1}$, and so on, and we conclude that $z_{m, i}$ is the image of $z_{m_i, i}$ under $A_{m_i} \to A_m$. We conclude that $z_{m, i}$ is congruent to $y_i$ modulo $A_{m, m_i + 1}$. In particular $z_{m, i} \not = 0$. This is impossible as $\sum z_{m, i} \in \bigoplus_i A_m$ hence only a finite number of $z_{m, i}$ can be nonzero.
It follows from Lemma \ref{lemma-Rlim-pro-equal} that (1) implies (2). Assume (2). Then $(A_n)$ is ML by Lemma \ref{lemma-emmanouil}. For $m \geq n$ let $A_{n, m} = \Im(A_m \to A_n)$ so that $A_n = A_{n, n} \supset A_{n, n + 1} \supset \ldots$. Note that $(A_n)$ is zero as a pro-object if and only if for every $n$ there is an $m \geq n$ such that $A_{n, m} = 0$. Note that $(A_n)$ is ML if and only if for every $n$ there is an $m_n \geq n$ such that $A_{n, m} = A_{n, m + 1} = \ldots$. In the ML case it is clear that $\lim A_n = 0$ implies that $A_{n, m_n} = 0$ because the maps $A_{n + 1, m_{n + 1}} \to A_{n, m}$ are surjective. This finishes the proof.
Let $K$ be represented by the system $(K_n^\bullet)$. Set $M_1^\bullet = K_1^\bullet$. Suppose we have constructed surjective maps of complexes $M_n^\bullet \to M_{n - 1}^\bullet \to \ldots \to M_1^\bullet$ and homotopy equivalences $\psi_e : K_e^\bullet \to M_e^\bullet$ such that the diagrams $$ \xymatrix{ K_{e + 1}^\bullet \ar[d] \ar[r] & K_e^\bullet \ar[d] \\ M_{e + 1}^\bullet \ar[r] & M_e^\bullet } $$ commute for all $e < n$. Then we consider the diagram $$ \xymatrix{ K_{n + 1}^\bullet \ar[r] & K_n^\bullet \ar[d] \\ & M_n^\bullet } $$ By Derived Categories, Lemma \ref{derived-lemma-make-surjective} we can factor the composition $K_{n + 1}^\bullet \to M_n^\bullet$ as $K_{n + 1}^\bullet \to M_{n + 1}^\bullet \to M_n^\bullet$ such that the first arrow is a homotopy equivalence and the second a termwise split surjection. The lemma follows from this and induction.
First use Lemma \ref{lemma-get-ML-system} to represent $K$ by a system of complexes $(M_n^\bullet)$ such that all the transition maps $M_{n + 1}^\bullet \to M_n^\bullet$ are surjective. Next, let $K_1^\bullet \to M_1^\bullet$ be a quasi-isomorphism with $K_1^\bullet$ a K-flat complex of $A_1$-modules (Lemma \ref{lemma-K-flat-resolution}). Suppose we have constructed $K_n^\bullet \to K_{n - 1}^\bullet \to \ldots \to K_1^\bullet$ and maps of complexes $\psi_e : K_e^\bullet \to M_e^\bullet$ such that $$ \xymatrix{ K_{e + 1}^\bullet \ar[d] \ar[r] & K_e^\bullet \ar[d] \\ M_{e + 1}^\bullet \ar[r] & M_e^\bullet } $$ commutes for all $e < n$. Then we consider the diagram $$ \xymatrix{ C^\bullet \ar@{..>}[d] \ar@{..>}[r] & K_n^\bullet \ar[d]^{\psi_n} \\ M_{n + 1}^\bullet \ar[r]^{\varphi_n} & M_n^\bullet } $$ in $D(A_{n + 1})$. As $M_{n + 1}^\bullet \to M_n^\bullet$ is termwise surjective, the complex $C^\bullet$ fitting into the left upper corner with terms $$ C^p = M_{n + 1}^p \times_{M_n^p} K_n^p $$ is quasi-isomorphic to $M_{n + 1}^\bullet$ (details omitted). Choose a quasi-isomorphism $K_{n + 1}^\bullet \to C^\bullet$ with $K_{n +1}^\bullet$ K-flat. Thus the lemma holds by induction.
Choose a representative $(K_n^\bullet)$ for $K$ such that each $K_n^\bullet$ is a K-flat complex (Lemma \ref{lemma-get-K-flat-system}). Then you can define $K \otimes^\mathbf{L} L$ as the object represented by the system of complexes $$ (\text{Tot}(K_n^\bullet \otimes_{A_n} L_n^\bullet)) $$ for any choice of representative $(L_n^\bullet)$ for $L$. This is well defined in both variables by Lemmas \ref{lemma-K-flat-quasi-isomorphism} and \ref{lemma-derived-tor-quasi-isomorphism-other-side}. Compatibility with the map to $D(A_n)$ is clear. Exactness follows exactly as in Lemma \ref{lemma-derived-tor-exact}.
Let $m \in M$. If $m$ maps to an element of $(M/M[I^\infty])[I]$ then $Im \subset M[I^\infty]$. Write $I = (f_1, \ldots, f_t)$. Then we see that $f_i m \in M[I^\infty]$, i.e., $I^{n_i}f_i m = 0$ for some $n_i > 0$. Thus we see that $I^Nm = 0$ with $N = \sum n_i + 2$. Hence $m$ maps to zero in $(M/M[I^\infty])$ which proves the first statement of the lemma. \medskip\noindent For the second, suppose that $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of modules with $M'$ and $M''$ both $I$-power torsion modules. Then $M[I^\infty] \supset M'$ and hence $M/M[I^\infty]$ is a quotient of $M''$ and therefore $I$-power torsion. Combined with the first statement and Lemma \ref{lemma-torsion-free} this implies that it is zero
It is clear that a submodule and a quotient module of an $I$-power torsion module is $I$-power torsion. Moreover, the extension of two $I$-power torsion modules is $I$-power torsion by Lemma \ref{lemma-divide-by-torsion}. Hence the statement of the lemma by Homology, Lemma \ref{homology-lemma-characterize-serre-subcategory}.
Proof of (2). We can write $N = \bigcup N[I^n]$. We have $K \otimes_R^\mathbf{L} N = \text{hocolim}_n K \otimes_R^\mathbf{L} N[I^n]$ as tensor products commute with colimits (details omitted; hint: represent $K$ by a K-flat complex and compute directly). Hence we may assume $N$ is annihilated by $I^n$. Consider the $R$-algebra $R' = R/I^n \oplus N$ where $N$ is an ideal of square zero. It suffices to show that $K' = K \otimes_R^\mathbf{L} R'$ is $0$ in $D(R')$. We have a surjection $R' \to R/I$ of $R$-algebras whose kernel $J$ is nilpotent (any product of $n$ elements in the kernel is zero). We have $$ 0 = K \otimes_R^\mathbf{L} R/I = (K \otimes_R^\mathbf{L} R') \otimes_{R'}^\mathbf{L} R/I = K' \otimes_{R'}^\mathbf{L} R/I $$ by Lemma \ref{lemma-double-base-change}. Hence by Lemma \ref{lemma-perfect-modulo-nilpotent-ideal} we find that $K'$ is a perfect complex of $R'$-modules. In particular $K'$ is bounded above and if $H^b(K')$ is the right-most nonvanishing cohomology module (if it exists), then $H^b(K')$ is a finite $R'$-module (use Lemmas \ref{lemma-perfect} and \ref{lemma-finite-cohomology}) with $H^b(K') \otimes_{R'} R'/J = H^b(K')/JH^b(K') = 0$ (because $K' \otimes_{R'}^\mathbf{L} R'/J = 0$). By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) we find $H^b(K') = 0$, i.e., $K' = 0$ as desired. \medskip\noindent Part (1) follows trivially from part (2). Part (3) follows from part (2), induction on the number of nonzero cohomology modules of $M$, and the distinguished triangles of truncation from Derived Categories, Remark \ref{derived-remark-truncation-distinguished-triangle}. Details omitted.
If $R \to S$ is flat, then $R/I^n \to S/I^nS$ is flat for every $n$, see Algebra, Lemma \ref{algebra-lemma-flat-base-change}. Hence (1) and (2) are equivalent by Algebra, Lemma \ref{algebra-lemma-ff-rings}. The equivalence of (1) with (3) follows by identifying $I$-torsion $R$-modules with $R/I$-modules, using that $$ M \otimes_R S = M \otimes_{R/I} S/IS $$ for $R$-modules $M$ annihilated by $I$, and Algebra, Lemma \ref{algebra-lemma-easy-ff}. The implication (4) $\Rightarrow$ (3) is immediate. Assume (3). We have seen above that $R/I^n \to S/I^nS$ is flat, and by assumption it induces a surjection on spectra, as $\Spec(R/I^n) = \Spec(R/I)$ and similarly for $S$. Hence the base change functor is faithful on modules annihilated by $I^n$. Since any $I$-power torsion module $M$ is the union $M = \bigcup M_n$ where $M_n$ is annihilated by $I^n$ we see that the base change functor is faithful on the category of all $I$-power torsion modules (as tensor product commutes with colimits).
Note that the equivalent conditions of both Lemma \ref{lemma-characterize-flatness-on-torsion} and Lemma \ref{lemma-neighbourhood-isomorphism} are satisfied. We will use these without further mention. We first prove (1). Let $M$ be any $R$-module. Set $M' = M/M[I^\infty]$ and consider the exact sequence $$ 0 \to M[I^\infty] \to M \to M' \to 0 $$ As $M[I^\infty] = M[I^\infty] \otimes_R S$ we see that it suffices to show that $(M' \otimes_R S)[(IS)^\infty] = 0$. Write $I = (f_1, \ldots, f_t)$. By Lemma \ref{lemma-divide-by-torsion} we see that $M'[I^\infty] = 0$. Hence for every $n > 0$ the map $$ M' \longrightarrow \bigoplus\nolimits_{i = 1, \ldots t} M', \quad x \longmapsto (f_1^n x, \ldots, f_t^n x) $$ is injective. As $S$ is flat over $R$ also the corresponding map $M' \otimes_R S \to \bigoplus_{i = 1, \ldots t} M' \otimes_R S$ is injective. This means that $(M' \otimes_R S)[I^n] = 0$ as desired. \medskip\noindent Next we prove (2). If $N$ is $I$-power torsion, then $N \otimes_R S = N$ and the displayed map of (2) is an isomorphism by Algebra, Lemma \ref{algebra-lemma-adjoint-tensor-restrict}. If $M$ is $I$-power torsion, then the image of any map $M \to N$ factors through $M[I^\infty]$ and the image of any map $M \otimes_R S \to N \otimes_R S$ factors through $(N \otimes_R S)[(IS)^\infty]$. Hence in this case part (1) guarantees that we may replace $N$ by $N[I^\infty]$ and the result follows from the case where $N$ is $I$-power torsion we just discussed. \medskip\noindent Next we prove (3). The functor is fully faithful by (2). For essential surjectivity, we simply note that for any $IS$-power torsion $S$-module $N$, the natural map $N \otimes_R S \to N$ is an isomorphism.
In both cases we have a complex $K_\bullet$ of $R$ modules and we want to show that $K_\bullet \to K_\bullet \otimes_R S$ is a quasi-isomorphism. By Lemma \ref{lemma-neighbourhood-isomorphism} and the flatness of $R \to S$ this will hold as soon as all homology groups of $K$ are $I$-power torsion. This is true for the Koszul complex by Lemma \ref{lemma-homotopy-koszul} and for the extended alternating {\v C}ech complex by Lemma \ref{lemma-extended-alternating-torsion}.
The notation above indicates the $\Ext$-groups in $\text{Mod}_R$ as defined in Homology, Section \ref{homology-section-extensions}. These are denoted $\Ext_R(M, N)$. Using the long exact sequence of Homology, Lemma \ref{homology-lemma-six-term-sequence-ext} associated to the short exact sequence $0 \to I \to R \to R/I \to 0$ and the fact that $\Ext_R(R, N) = 0$ we see that $$ \Ext_R(R/I, N) = \Coker(N \longrightarrow \Hom(I, N)) $$ Using the short exact sequence of Lemma \ref{lemma-naive-Koszul-complex} we see that we get a complex $$ N \to \Hom(M, N) \to \Hom_R(K, N) $$ whose homology in the middle is canonically isomorphic to $\Ext_R(R/I, N)$. The proof of the lemma is now complete as the cokernel of the first map is canonically isomorphic to $H_1(N, f_\bullet)$.
As $M$ is $I$-power torsion we see that $M \otimes_R S = M$, see Lemma \ref{lemma-neighbourhood-isomorphism}. We will use this identification without further mention. As $R \to S$ is flat, the base change functor is exact and we obtain a functorial map of $\Ext$-groups $$ \Ext_R(M, N) \longrightarrow \Ext_S(M \otimes_R S, N \otimes_R S), $$ see Homology, Lemma \ref{homology-lemma-exact-functor-ext}. The claim of the lemma is that this map is surjective when $M$ is $I$-power torsion. In fact we will show that it is an isomorphism. By Lemma \ref{lemma-I-power-torsion-presentation} we can find a surjection $M' \to M$ with $M'$ a direct sum of modules of the form $R/I^n$. Using the long exact sequence of Homology, Lemma \ref{homology-lemma-six-term-sequence-ext} we see that it suffices to prove the lemma for $M'$. Using compatibility of $\Ext$ with direct sums (details omitted) we reduce to the case where $M = R/I^n$ for some $n$. \medskip\noindent Let $f_1, \ldots, f_t$ be generators for $I^n$. By Lemma \ref{lemma-explicit-ext} we have a commutative diagram $$ \xymatrix{ 0 \ar[r] & \Ext_R(R/I^n, N) \ar[r] \ar[d] & H_1(N, f_\bullet) \ar[r] \ar[d] & \Hom_R(K, N) \ar[d] \\ 0 \ar[r] & \Ext_S(S/I^nS, N \otimes S) \ar[r] & H_1(N \otimes S, f_\bullet) \ar[r] & \Hom_S(K \otimes S, N \otimes S) } $$ with exact rows where $K$ is as in Lemma \ref{lemma-naive-Koszul-complex}. Hence it suffices to prove that the two right vertical arrows are isomorphisms. Since $K$ is annihilated by $I^n$ we see that $\Hom_R(K, N) = \Hom_S(K \otimes_R S, N \otimes_R S)$ by Lemma \ref{lemma-neighbourhood-equivalence}. As $R \to S$ is flat we have $H_1(N, f_\bullet) \otimes_R S = H_1(N \otimes_R S, f_\bullet)$. As $H_1(N, f_\bullet)$ is annihilated by $I^n$, see Lemma \ref{lemma-koszul-homology-annihilated} we have $H_1(N, f_\bullet) \otimes_R S = H_1(N, f_\bullet)$ by Lemma \ref{lemma-neighbourhood-isomorphism}.
First proof. Denote $\check{\mathcal{C}}_R \to \check{\mathcal{C}}_S$ the quasi-isomorphism of extended alternating {\v C}ech complexes of Lemma \ref{lemma-map-identifies-koszul-and-cech-complexes}. Since these complexes are bounded with flat terms, we see that $M \otimes_R \check{\mathcal{C}}_R \to M \otimes_R \check{\mathcal{C}}_S$ is a quasi-isomorphism too (Lemmas \ref{lemma-derived-tor-quasi-isomorphism} and \ref{lemma-derived-tor-quasi-isomorphism-other-side}). Now the complex (\ref{equation-glueing-complex}) is a truncation of the cone of the map $M \otimes_R \check{\mathcal{C}}_R \to M \otimes_R \check{\mathcal{C}}_S$ and we win. \medskip\noindent Second computational proof. Let $m \in M$. If $\alpha(m) = 0$, then $m \in M[I^\infty]$, see Lemma \ref{lemma-torsion-free}. Pick $n$ such that $I^n m = 0$ and consider the map $\varphi : R/I^n \to M$. If $m \otimes 1 = 0$, then $\varphi \otimes 1_S = 0$, hence $\varphi = 0$ (see Lemma \ref{lemma-neighbourhood-equivalence}) hence $m = 0$. In this way we see that $\alpha$ is injective. \medskip\noindent Let $(m', m'_1, \ldots, m'_t) \in \Ker(\beta)$. Write $m'_i = m_i/f_i^n$ for some $n > 0$ and $m_i \in M$. We may, after possibly enlarging $n$ assume that $f_i^n m' = m_i \otimes 1$ in $M \otimes_R S$ and $f_j^nm_i - f_i^nm_j = 0$ in $M$. In particular we see that $(m_1, \ldots, m_t)$ defines an element $\xi$ of $H_1(M, (f_1^n, \ldots, f_t^n))$. Since $H_1(M, (f_1^n, \ldots, f_t^n))$ is annihilated by $I^{tn + 1}$ (see Lemma \ref{lemma-koszul-homology-annihilated}) and since $R \to S$ is flat we see that $$ H_1(M, (f_1^n, \ldots, f_t^n)) = H_1(M, (f_1^n, \ldots, f_t^n)) \otimes_R S = H_1(M \otimes_R S, (f_1^n, \ldots, f_t^n)) $$ by Lemma \ref{lemma-neighbourhood-isomorphism} The existence of $m'$ implies that $\xi$ maps to zero in the last group, i.e., the element $\xi$ is zero. Thus there exists an $m \in M$ such that $m_i = f_i^n m$. Then $(m', m'_1, \ldots, m'_t) - \alpha(m) = (m'', 0, \ldots, 0)$ for some $m'' \in (M \otimes_R S)[(IS)^\infty]$. By Lemma \ref{lemma-neighbourhood-equivalence} we conclude that $m'' \in M[I^\infty]$ and we win.
Consider an object $\mathbf{M} = (M', M_i, \alpha_i, \alpha_{ij})$ of $\text{Glue}(R \to S, f_1, \ldots, f_t)$. By Algebra, Lemma \ref{algebra-lemma-glue-modules} there exists a unique module $M$ and isomorphisms $M_{f_i} \to M_i$ which recover the glueing data $\alpha_{ij}$. Then both $M'$ and $M \otimes_R S$ are $S$-modules which recover the modules $M_i \otimes_R S$ upon localizing at $f_i$. Whence there is a canonical isomorphism $M \otimes_R S \to M'$. This shows that $\mathbf{M}$ is in the essential image of $\text{Can}$. Combined with Lemma \ref{lemma-H0-inverse} the lemma follows.