hcju/stacksqa huggingface dataset:
By Lemma \ref{lemma-derived-completion} we know that derived completion is given by $R\Hom_A(C, -)$ for some $C \in D^b(A)$. By Lemmas \ref{lemma-derived-completion-finite-cohomological-dimension} and \ref{lemma-projective-amplitude} we see that $C$ has finite projective dimension. Thus we may choose a bounded complex of projective modules $P^\bullet$ representing $C$. Then $$ M^\bullet = \Hom^\bullet(P^\bullet, K^\bullet) $$ is a complex of $A$-modules representing $(K^\bullet)^\wedge$. It comes with a filtration given by $F^pM^\bullet = \Hom^\bullet(P^\bullet, F^pK^\bullet)$. We see that $F^pM^\bullet$ represents $(F^pK^\bullet)^\wedge$ and hence $\text{gr}^pM^\bullet$ represents $(\text{gr}K^\bullet)^\wedge$. Thus we find our spectral sequence by taking the spectral sequence of the filtered complex $M^\bullet$, see Homology, Section \ref{homology-section-filtered-complex}. If the filtration on each $K^n$ is finite, then the filtration on each $M^n$ is finite because $P^\bullet$ is a bounded complex. Hence the final statement follows from Homology, Lemma \ref{homology-lemma-biregular-ss-converges}.
Using Lemma \ref{lemma-ideal-of-elements-complete-wrt} we see that $L \in D(B)$ is in $D_{comp}(B, IB)$ if and only if $T(L, f)$ is zero for every local section $f \in I$. Observe that the cohomology of $T(L, f)$ is computed in the category of abelian groups, so it doesn't matter whether we think of $f$ as an element of $A$ or take the image of $f$ in $B$. The lemma follows immediately from this and the definition of derived complete objects.
Choose generators $f_1, \ldots, f_r$ of $I$. Denote $\check{\mathcal{C}}^\bullet_A \to \check{\mathcal{C}}^\bullet_B$ the quasi-isomorphism of extended alternating {\v C}ech complexes of Lemma \ref{lemma-map-identifies-koszul-and-cech-complexes}. Let $K \in D_{comp}(A, I)$. Let $I^\bullet$ be a K-injective complex of $A$-modules representing $K$. Since $\Ext^n_A(A_f, K)$ and $\Ext^n_A(B_f, K)$ are zero for all $f \in I$ and $n \in \mathbf{Z}$ (Lemma \ref{lemma-hom-from-Af}) we conclude that $\check{\mathcal{C}}^\bullet_A \to A$ and $\check{\mathcal{C}}^\bullet_B \to B$ induce quasi-isomorphisms $$ I^\bullet = \Hom_A(A, I^\bullet) \longrightarrow \text{Tot}(\Hom_A(\check{\mathcal{C}}^\bullet_A, I^\bullet)) $$ and $$ \Hom_A(B, I^\bullet) \longrightarrow \text{Tot}(\Hom_A(\check{\mathcal{C}}^\bullet_B, I^\bullet)) $$ Some details omitted. Since $\check{\mathcal{C}}^\bullet_A \to \check{\mathcal{C}}^\bullet_B$ is a quasi-isomorphism and $I^\bullet$ is K-injective we conclude that $\Hom_A(B, I^\bullet) \to I^\bullet$ is a quasi-isomorphism. As the complex $\Hom_A(B, I^\bullet)$ is a complex of $B$-modules we conclude that $K$ is in the image of the restriction map, i.e., the functor is essentially surjective \medskip\noindent In fact, the argument shows that $F : D_{comp}(A, I) \to D_{comp}(B, IB)$, $K \mapsto \Hom_A(B, I^\bullet)$ is a left inverse to restriction. Finally, suppose that $L \in D_{comp}(B, IB)$. Represent $L$ by a K-injective complex $J^\bullet$ of $B$-modules. Then $J^\bullet$ is also K-injective as a complex of $A$-modules (Lemma \ref{lemma-K-injective-flat}) hence $F(\text{restriction of }L) = \Hom_A(B, J^\bullet)$. There is a map $J^\bullet \to \Hom_A(B, J^\bullet)$ of complexes of $B$-modules, whose composition with $\Hom_A(B, J^\bullet) \to J^\bullet$ is the identity. We conclude that $F$ is also a right inverse to restriction and the proof is finished.
If the $f$-power torsion is bounded, then the pro-objects $\{(f^n : A \to A)\}$ and $\{A/f^nA\}$ are isomorphic by Lemma \ref{lemma-lift-universally}. Hence the functors are isomorphic by Lemma \ref{lemma-Rlim-pro-equal}. Conversely, we see from Lemma \ref{lemma-tensor-Rlim-exact} that the condition is exactly that $$ R\lim (K \otimes_A^\mathbf{L} A[f^n]) $$ is zero for all $K \in D(A)$. Here the maps of the system $(A[f^n])$ are given by multiplication by $f$. Taking $K = A$ and $K = \bigoplus_{i \in \mathbf{N}} A$ we see from Lemma \ref{lemma-Rlim-zero-of-direct-sums} this implies $(A[f^n])$ is zero as a pro-object, i.e., $f^{n - 1}A[f^n] = 0$ for some $n$, i.e., $A[f^{n - 1}] = A[f^n]$, i.e., the $f$-power torsion is bounded.
Say $I = (f_1, \ldots, f_r)$. It suffices to show that for each $i$ there is an $n_i$ such that $f_i^{n_i}M = 0$. Hence we may assume that $I = (f)$ is a principal ideal. Let $B = \mathbf{Z}[x] \to A$ be the ring map sending $x$ to $f$. By Lemma \ref{lemma-restriction-derived-complete} we see that $M$ is derived complete as a $B$-module with respect to the ideal $(x)$. After replacing $A$ by $B$, we may assume that $f$ is a nonzerodivisor in $A$. \medskip\noindent Assume $I = (f)$ with $f \in A$ a nonzerodivisor. According to Example \ref{example-derived-complete-modules} there exists a short exact sequence $$ 0 \to K \xrightarrow{u} L \to M \to 0 $$ where $K$ and $L$ are $I$-adically complete $A$-modules whose $f$-torsion is zero\footnote{For the proof it is enough to show that there exists a sequence $K \xrightarrow{u} L \to M \to 0$ where $K$ and $L$ are $I$-adically complete $A$-modules. This can be shown by choosing a presentation $F_1 \to F_0 \to M \to 0$ with $F_i$ free and then setting $K$ and $L$ equal to the $f$-adic completions of $F_1$ and $F_0$. Namely, as $f$ is a nonzerodivisor these completions will be the derived completions and the sequence will remain exact.}. Consider $K$ and $L$ as topological modules with the $I$-adic topology. Then $u$ is continuous. Let $$ L_n = \{x \in L \mid f^n x \in u(K)\} $$ Since $M$ is $f$-power torsion we see that $L = \bigcup L_n$. Let $N_n$ be the closure of $L_n$ in $L$. By Lemma \ref{lemma-consequence-baire-complete-module} we see that $N_n$ is open in $L$ for some $n$. Fix such an $n$. Since $f^{n + m} : L \to L$ is a continuous open map, and since $f^{n + m} L_n \subset u(f^m K)$ we conclude that the closure of $u(f^mK)$ is open for all $m \geq 1$. Thus by Lemma \ref{lemma-open-mapping} we conclude that $u$ is open. Hence $f^tL \subset \Im(u)$ for some $t$ and we conclude that $f^t$ annihilates $M$ as desired.
Take $x \in M'$ and $g \in J$. For every $n \geq 1$ we may write $x = f^n x_n$. Since $g$ is in $f^nA$ we see that the element $y_n = gx_n$ in $M'$ is independent of the choice of $x_n$. In particular, we may take $x_n = fx_{n + 1}$ and we find that $y_n = fy_{n + 1}$. Thus we obtain a map $A_f \to M$ sending $1/f^n$ to $y_n$. This map has to be zero as $M$ is derived complete (Lemma \ref{lemma-hom-from-Af}) and hence $y_n = 0$ for all $n$. Since $gx = gfx_1 = fy_1$ this completes the proof.
Let $f \in I$. By Lemma \ref{lemma-largest-ideal-henselian} it suffices to show that $(A, fA)$ is a henselian pair. Observe that $A$ is derived complete with respect to $fA$ (follows immediately from Definition \ref{definition-derived-complete}). By Lemma \ref{lemma-complete-derived-complete} the map from $A$ to the $f$-adic completion $A'$ of $A$ is surjective. By Lemma \ref{lemma-complete-henselian} the pair $(A', fA')$ is henselian. Thus it suffices to show that $(A, \bigcap f^nA)$ is a henselian pair, see Lemma \ref{lemma-henselian-henselian-pair}. This follows from Lemmas \ref{lemma-kernel-to-completion-square-zero} and \ref{lemma-locally-nilpotent-henselian}.
When $r = 1$ this is Lemma \ref{lemma-kernel-to-completion-square-zero}. Say $I = (f_1, \ldots, f_r)$ with $r > 1$. By Lemma \ref{lemma-serre-subcategory} the ring $A_t = A/f_r^tA$ is derived complete with respect to $I$ and hence a fortiori derived complete with respect to $I_t = (f_1, \ldots, f_{r - 1})A_t$. Observe that $A \to A_t$ sends $J$ into $J_t = \bigcap I_t^n$. By induction $J_t^{N/2} = 0$ with $N = 2^r$. The ideal $\bigcap \Ker(A \to A_t) = \bigcap f_r^t A$ has square zero by the case $r = 1$. This finishes the proof.
We have an inverse system of distinguished triangles $$ \tau_{\leq -1}K_n^\bullet \to K_n^\bullet \to A/(f_1^m, \ldots, f_r^m) \to (\tau_{\leq -1}K_n^\bullet)[1] $$ See Derived Categories, Remark \ref{derived-remark-truncation-distinguished-triangle}. By Derived Categories, Lemma \ref{derived-lemma-pro-isomorphism} it suffices to show that the inverse system $\tau_{\leq -1}K_n^\bullet$ is pro-zero. Recall that $K_n^\bullet$ has nonzero terms only in degrees $i$ with $-r \leq i \leq 0$. Thus by Derived Categories, Lemma \ref{derived-lemma-essentially-constant-cohomology} it suffices to show that $H^p(K_n^\bullet)$ is pro-zero for $p \leq -1$. In other words, for every $n \in \mathbf{N}$ we have to show there exists an $m \geq n$ such that $H^p(K_m^\bullet) \to H^p(K_n^\bullet)$ is zero. Since $A$ is Noetherian, we see that $$ H^p(K_n^\bullet) = \frac{\Ker(K_n^p \to K_n^{p + 1})}{\Im(K_n^{p - 1} \to K_n^p)} $$ is a finite $A$-module. Moreover, the map $K_m^p \to K_n^p$ is given by a diagonal matrix whose entries are in the ideal $(f_1^{m - n}, \ldots, f_r^{m - n})$ as $p < 0$. Note that $H^p(K_n^\bullet)$ is annihilated by $J = (f_1^n, \ldots, f_r^n)$, see Lemma \ref{lemma-homotopy-koszul}. Now $J^t \subset (f_1^{m - n}, \ldots, f_r^{m - n})$ for $m = n + tr$. Thus by Artin-Rees (Algebra, Lemma \ref{algebra-lemma-Artin-Rees}) for some $m$ large enough we see that the image of $K_m^p \to K_n^p$ intersected with $\Ker(K_n^p \to K_n^{p + 1})$ is contained in $J \Ker(K_n^p \to K_n^{p + 1})$. For this $m$ we get the zero map.
The complex $\tau_{\leq m}K$ is pseudo-coherent for all $m$ by Lemma \ref{lemma-Noetherian-pseudo-coherent}. Thus $\tau_{\leq m}K$ is represented by a bounded above complex $P^\bullet$ of finite free $A$-modules. Then $\tau_{\leq m}K \otimes_A^\mathbf{L} A/I^n = P^\bullet/I^nP^\bullet$. Hence $(\tau_{\leq m}K)^\wedge = R\lim P^\bullet/I^nP^\bullet$ (Proposition \ref{proposition-noetherian-naive-completion-is-completion}) and since the $R\lim$ is just given by termwise $\lim$ (Lemma \ref{lemma-compute-Rlim-modules}) and since $I$-adic completion is an exact functor on finite $A$-modules (Algebra, Lemma \ref{algebra-lemma-completion-flat}) we conclude the result holds for $\tau_{\leq m}K$. Hence the result holds for $K$ as derived completion has finite cohomological dimension, see Lemma \ref{lemma-derived-completion-finite-cohomological-dimension}.
Assume $M/IM$ is finite. Pick $x_1, \ldots, x_t \in M$ which map to generators of $M/IM$. We obtain a map $A^{\oplus t} \to M$ mapping the $i$th basis vector to $x_i$. By Proposition \ref{proposition-noetherian-naive-completion-is-completion} the derived completion of $A$ is $A^\wedge = \lim A/I^n$. As $M$ is derived complete, we see that our map factors through a map $q : (A^\wedge)^{\oplus t} \to M$. The module $\Coker(q)$ is zero by Lemma \ref{lemma-derived-complete-zero}. Thus $M$ is a finite $A^\wedge$-module. Since $A^\wedge$ is Noetherian and complete with respect to $IA^\wedge$, it follows that $M$ is $I$-adically complete (use Algebra, Lemmas \ref{algebra-lemma-completion-Noetherian}, \ref{algebra-lemma-when-finite-module-complete-over-complete-ring}, and \ref{algebra-lemma-Artin-Rees}).
For an $A$-module $M$ denote $M^\wedge$ the derived completion and $\lim M/I^nM$ the usual completion. Assume $M$ is finite. The system $\text{Tor}^A_i(M, A/I^n)$ is pro-zero for $i > 0$, see Lemma \ref{lemma-tor-strictly-pro-zero}. Since $\text{Tor}_i^A(M \otimes_A N, A/I^n) = \text{Tor}_i^A(M, A/I^n) \otimes_A N$ as $N$ is flat, the same is true for the system $\text{Tor}^A_i(M \otimes_A N, A/I^n)$. By Lemma \ref{lemma-noetherian-calculate} we conclude $R\lim (M \otimes_A N) \otimes_A^\mathbf{L} A/I^n$ only has cohomology in degree $0$ given by the usual completion $\lim M \otimes_A N/ I^n(M \otimes_A N)$. This proves (1). Part (2) follows from (1) and the fact that $M_f = M \otimes_A A_f$.
Let $L$ be as in (2). We may represent $K$ by a bounded above complex $P^\bullet$ of free $A$-modules. We may represent $L$ by a bounded above complex $F^\bullet$ of finite free $A$-modules. Since $\text{Tot}(P^\bullet \otimes_A F^\bullet)$ represents $K \otimes_A^\mathbf{L} L$ we see that $(K \otimes_A^\mathbf{L} L)^\wedge$ is represented by $$ \text{Tot}((P^\bullet)^\wedge \otimes_A F^\bullet) $$ where $(P^\bullet)^\wedge$ is the complex whose terms are the usual $=$ derived completions $(P^n)^\wedge$, see for example Proposition \ref{proposition-noetherian-naive-completion-is-completion} and Lemma \ref{lemma-when-derived-completion-is-completion}. This proves (2). Part (1) is a special case of (2).
Let $x \in (\eta_fM)^i$. Then $x = f^ix' \in f^iM$ and $d^i(x) = f^{i + 1}y \in f^{i + 1}M^{i + 1}$. Thus $d^i$ maps $x' \otimes f^i$ to zero in $M^{i + 1} \otimes f^iA/f^{i + 1}A$. All tensor products are over $A$ in this proof. Hence we may map $x$ to the class of $x' \otimes f^i$ in $H^i(M^\bullet \otimes f^iA/f^{i + 1}A)$. It is clear that this rule defines a map $$ (\eta_fM)^i \otimes A/fA \longrightarrow H^i(M^\bullet \otimes f^iA/f^{i + 1}A) $$ of $A/fA$-modules. Observe that in the situation above, we may view $x' \otimes f^i$ as an element of $M^i \otimes f^iA/f^{i + 2}A$ with differential $d^i(x' \otimes f^i) = y \otimes f^{i + 1}$. By the construction of $\beta$ above we find that $\beta(x' \otimes f^i) = y \otimes f^{i + 1}$ and we conclude that our maps are compatible with differentials, i.e., we have a map of complexes. \medskip\noindent To finish the proof, we observe that the construction given in the previous paragraph agrees with the maps $(\eta_fM)^i \otimes A/fA \to Z^i/B^i$ discussed in Remark \ref{remark-eta-BZ}. Since we have seen that the kernel of these maps is an acyclic subcomplex of $\eta_fM^\bullet \otimes A/fA$, the lemma is proved.
Observe that $I$ contains the element $f^{r_i}$. Hence $I = (g)$ where $g$ divides a power of $f$. Consider the complex of finite free $A$-modules $$ 0 \to M^i \xrightarrow{f, d^i} M^i \oplus M^{i + 1} \xrightarrow{d^i, -f} M^{i + 1} \to 0 $$ which becomes split exact after localizing at $f$. By construction $(\eta_fM)^i$ is isomorphic to the kernel of the second map. Hence $(\eta_fM)^i$ is the kernel of the map $$ M^i \oplus M^{i + 1} \longrightarrow Q = \Coker(f, d^i)/f\text{-power torsion} $$ By Lemma \ref{lemma-principal-fitting-ideal} the finite module $Q$ can locally be generated by $r_{i + 1}$ elements. On the other hand we have $Q_f \cong M^{i + 1}_f$ is free of rank $r_{i + 1}$. Hence $Q$ is locally free of rank $r_{i + 1}$. Thus $Q$ is a projective module and the surjection above is split.
Set $N^i = M^i \otimes_A B$. Observe that $f^iM^i \otimes_A B = g^iN^i$ as submodules of $(N^i)_g$. The maps $$ (\eta_fM)^i \otimes_A B \to g^iN^i \otimes g^{i + 1}N^{i + 1} \quad\text{and}\quad (\eta_gN)^i \to g^iN^i \otimes g^{i + 1}N^{i + 1} $$ are inclusions of direct summands by Lemma \ref{lemma-eta-locally-free}. Since their images agree after localizing at $g$ we conclude.
With notation as in Remark \ref{remark-eta-BZ} we define a map $t^{-1} : Z^i \to (\eta_fM)^i / f(\eta_fM)^i$. Namely, for $x \in M^i$ with $d^i(x) = f^2y$ we send the class of $x$ in $Z^i$ to the class of $f^ix$ in $(\eta_fM)^i / f(\eta_fM)^i$. We omit the verification that this is well defined; the assumption of the lemma exactly signifies that the domain of this operation is all of $Z^i$. Then $t \circ t^{-1} = \text{id}_{Z^i}$. Hence $t^{-1}$ defines a splitting of the short exact sequence in Remark \ref{remark-eta-BZ} and the resulting direct sum decomposition $$ (\eta_fM)^i / f(\eta_fM)^i = Z^i \oplus B^{i + 1} $$ is compatible with the map displayed in the lemma.
By Lemma \ref{lemma-eta-locally-free} we find that $(\eta_fM)^i$ is free of rank $r_i$ for all $i$. Consider the map $$ (\eta_fM)^i / f(\eta_fM)^i \longrightarrow f^iM^i/f^{i + 1}M^i \oplus f^{i + 1}M^{i + 1}/f^{i + 2}M^{i + 1} $$ of Lemma \ref{lemma-eta-vanishing-beta}. By Lemma \ref{lemma-eta-locally-free} this map is split injective. Let $I_i \subset A$ be the ideal containing $f$ such that the ideal $I_i/fA \subset A/fA$ is the one found in Lemma \ref{lemma-ideal-direct-summand} for the displayed arrow. We set $I = \sum_{i \in \mathbf{Z}} I_i$. Since almost all $I_i = A$ this is a finitely generated ideal. \medskip\noindent Pick $\mathfrak p \in E$. By Lemma \ref{lemma-eta-vanishing-beta} and the freeness of the modules $(\eta_fM)^i$ we may write $$ \left((\eta_fM)^i / f(\eta_fM)^i\right)_\mathfrak p = (A/fA)_\mathfrak p^{\oplus m_i} \oplus (A/fA)_\mathfrak p^{\oplus n_i} $$ compatible with the arrow above. By the universal property of the ideal $I_i$ we conclude that $(I_i)_\mathfrak p = fA_\mathfrak p$. Hence $I_\mathfrak p = fA_\mathfrak p$ for $\mathfrak p \in E$. \medskip\noindent To finish the proof we need to show the assertion about cohomology. To see this, observe that the differential on $\eta_fM^\bullet$ fits into a commutative diagram $$ \xymatrix{ (\eta_fM)^i \ar[d] \ar[r] & f^iM^i \oplus f^{i + 1}M^{i + 1} \ar[d]^{\left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)} \\ (\eta_fM)^{i + 1} \ar[r] & f^{i + 1}M^i \oplus f^{i + 2}M^{i + 2} } $$ By construction, after tensoring with $A/I$, the modules on the left are direct sums of direct summands of the summands on the right. Picture $$ \xymatrix{ (\eta_fM)^i \otimes A/I \ar[d] \ar@{=}[r] & K^i \oplus L^i \ar[r] \ar[d] & f^iM^i \otimes A/I \oplus f^{i + 1}M^{i + 1} \otimes A/I \ar[d]^{\left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)} \\ (\eta_fM)^{i + 1} \otimes A/I \ar@{=}[r] & K^{i + 1} \oplus L^{i + 1} \ar[r] & f^{i + 1}M^i \otimes A/I \oplus f^{i + 2}M^{i + 2} \otimes A/I } $$ where the horizontal arrows are compatible with direct sum decompositions as well as inclusions of direct summands. It follows that the differential identifies $L^i$ with a direct summand of $K^{i + 1}$ and we conclude that the cohomology of $\eta_fM^\bullet \otimes_A A/I$ in degree $i$ is the module $K^{i + 1}/L^i$ which is finite projective as desired.
By assumption we can find a bounded above complex of finite free $A_1$-modules $P_1^\bullet$ representing $K_1$, see Definition \ref{definition-pseudo-coherent}. By Lemma \ref{lemma-lift-complex-stably-frees} we can, by induction on $n > 1$, find complexes $P_n^\bullet$ of finite free $A_n$-modules representing $K_n$ and maps $P_n^\bullet \to P_{n - 1}^\bullet$ representing the maps $K_n \to K_{n - 1}$ inducing isomorphisms (!) of complexes $P_n^\bullet \otimes_{A_n} A_{n - 1} \to P_{n - 1}^\bullet$. Thus $K = R\lim K_n$ is represented by $P^\bullet = \lim P_n^\bullet$, see Lemma \ref{lemma-compute-Rlim-modules} and Remark \ref{remark-how-unique}. Since $P_n^i$ is a finite free $A_n$-module for each $n$ and $A = \lim A_n$ we see that $P^i$ is finite free of the same rank as $P_1^i$ for each $i$. This means that $K$ is pseudo-coherent. It also follows that $K \otimes_A^\mathbf{L} A_n$ is represented by $P^\bullet \otimes_A A_n = P_n^\bullet$ which proves the final assertion.
We already know that $K$ is pseudo-coherent and that $K \otimes_A^\mathbf{L} A/I^n \to K_n$ is an isomorphism for all $n$, see Lemma \ref{lemma-Rlim-pseudo-coherent-gives-pseudo-coherent}. To finish the proof it suffices to show that $K$ is derived complete. This follows from Lemma \ref{lemma-pseudo-coherent-is-derived-complete}.
We already know that $K$ is pseudo-coherent and that $K \otimes_A^\mathbf{L} A_n \to K_n$ is an isomorphism for all $n$ by Lemma \ref{lemma-Rlim-pseudo-coherent-gives-pseudo-coherent}. Thus it suffices to show that $H^i(K \otimes_A^\mathbf{L} \kappa) = 0$ for $i \ll 0$ and every surjective map $A \to \kappa$ whose kernel is a maximal ideal $\mathfrak m$, see Lemma \ref{lemma-check-perfect-stalks}. Any element of $A$ which maps to a unit in $A_1$ is a unit in $A$ by Algebra, Lemma \ref{algebra-lemma-locally-nilpotent-unit} and hence $\Ker(A \to A_1)$ is contained in the Jacobson radical of $A$ by Algebra, Lemma \ref{algebra-lemma-contained-in-radical}. Hence $A \to \kappa$ factors as $A \to A_1 \to \kappa$. Hence $$ K \otimes_A^\mathbf{L} \kappa = K \otimes_A^\mathbf{L} A_1 \otimes_{A_1}^\mathbf{L} \kappa = K_1 \otimes_{A_1}^\mathbf{L} \kappa $$ and we get what we want as $K_1$ has finite tor dimension by Lemma \ref{lemma-perfect}.
Suppose that $H^i(K_1) = 0$ for $i > b$. Then we can find a complex of free $A/I$-modules $P_1^\bullet$ representing $K_1$ with $P_1^i = 0$ for $i > b$. By Lemma \ref{lemma-lift-complex-projectives} we can, by induction on $n > 1$, find complexes $P_n^\bullet$ of free $A/I^n$-modules representing $K_n$ and maps $P_n^\bullet \to P_{n - 1}^\bullet$ representing the maps $K_n \to K_{n - 1}$ inducing isomorphisms (!) of complexes $P_n^\bullet/I^{n - 1}P_n^\bullet \to P_{n - 1}^\bullet$. \medskip\noindent Thus we have arrived at the situation where $R\lim K_n$ is represented by $P^\bullet = \lim P_n^\bullet$, see Lemma \ref{lemma-compute-Rlim-modules} and Remark \ref{remark-how-unique}. The complexes $P_n^\bullet$ are uniformly bounded above complexes of flat $A/I^n$-modules and the transition maps are termwise surjective. Then $P^\bullet$ is a bounded above complex of flat $A$-modules by Lemma \ref{lemma-limit-flat}. It follows that $K \otimes_A^\mathbf{L} A/I^t$ is represented by $P^\bullet \otimes_A A/I^t$. We have $P^\bullet \otimes_A A/I^t = \lim P_n^\bullet \otimes_A A/I^t$ termwise by Lemma \ref{lemma-limit-flat}. The transition maps $P_{n + 1}^\bullet \otimes_A A/I^t \to P_n^\bullet \otimes_A A/I^t$ are isomorphisms for $n \geq t$. Hence we have $\lim P_n^\bullet \otimes_A A/I^t = R\lim P_n^\bullet \otimes_A A/I^t$. By assumption and our choice of $P_n^\bullet$ the complex $P_n^\bullet \otimes_A A/I^t = P_n^\bullet \otimes_{A/I^n} A/I^t$ represents $K_n \otimes_{A/I^n}^\mathbf{L} A/I^t = K_t$ for all $n \geq t$. We conclude $$ P^\bullet \otimes_A A/I^t = R\lim P_n^\bullet \otimes_A A/I^t = R\lim K_t = K_t $$ In other words, we have $K \otimes_A^\mathbf{L} A/I^t = K_t$. This proves the lemma as it follows that $K$ is derived complete by Proposition \ref{proposition-noetherian-naive-completion-is-completion}.
Recall that $K^\wedge = R\lim K_n$ is the derived completion of $K$, see Proposition \ref{proposition-noetherian-naive-completion-is-completion}. By Lemma \ref{lemma-derived-completion-pseudo-coherent} we have $H^i(K^\wedge) = \lim H^i(K)/I^nH^i(K)$. By Lemma \ref{lemma-break-long-exact-sequence-modules} we get short exact sequences $$ 0 \to R^1\lim H^{i - 1}(K_n) \to H^i(K^\wedge) \to \lim H^i(K_n) \to 0 $$ The Mittag-Leffler condition guarantees that the left terms are zero (Lemma \ref{lemma-compute-Rlim-modules}) and we conclude the lemma is true.
Choose a K-injective complex $I^\bullet$ representing $M$, a K-injective complex $J^\bullet$ representing $L$, and a bounded above complex of finite projective modules $K^\bullet$ representing $K$. Consider the map of complexes $$ \text{Tot}(\Hom^\bullet(J^\bullet, I^\bullet) \otimes_R K^\bullet) \longrightarrow \Hom^\bullet(\Hom^\bullet(K^\bullet, J^\bullet), I^\bullet) $$ of Lemma \ref{lemma-evaluate-and-more}. Note that $$ \left(\prod\nolimits_{p + r = t} \Hom_R(J^{-r}, I^p)\right) \otimes_R K^s = \prod\nolimits_{p + r = t} \Hom_R(J^{-r}, I^p) \otimes_R K^s $$ because $K^s$ is finite projective. The map is given by the maps $$ c_{p, r, s} : \Hom_R(J^{-r}, I^p) \otimes_R K^s \longrightarrow \Hom_R(\Hom_R(K^s, J^{-r}), I^p) $$ which are isomorphisms as $K^s$ is finite projective. For every element $\alpha = (\alpha^{p, r, s})$ of degree $n$ of the left hand side, there are only finitely many values of $s$ such that $\alpha^{p, r, s}$ is nonzero (for some $p, r$ with $n = p + r + s$). Hence our map is an isomorphism if the same vanishing condition is forced on the elements $\beta = (\beta^{p, r, s})$ of the right hand side. If $K^\bullet$ is a bounded complex of finite projective modules, this is clear. On the other hand, if we can choose $I^\bullet$ bounded and $J^\bullet$ bounded below, then $\beta^{p, r, s}$ is zero for $p$ outside a fixed range, for $s \gg 0$, and for $r \gg 0$. Hence among solutions of $n = p + r + s$ with $\beta^{p, r, s}$ nonzero only a finite number of $s$ values occur.