hcju/stacksqa huggingface dataset:
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 \ref{remark-glueing-data}. By Proposition \ref{proposition-equivalence} 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 $$ (M \otimes_R S, M_{f_i}, can_i, can_{ij}) = (M', N_{f_i}, \varphi_{f_i}, can_{ij}) \to (N \otimes_R S, N_{f_i}, can_i, can_{ij}) $$ 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 \ref{proposition-equivalence}). 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 $\varphi$. \medskip\noindent 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 \ref{remark-glueing-data}. By Proposition \ref{proposition-equivalence} 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 $$ (M \otimes_R S, M_{f_i}, can_i, can_{ij}) \to (N', M_{f_i}, \varphi_{f_i}, can_{ij}) = (N \otimes_R S, N_{f_i}, can_i, can_{ij}) $$ 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 \ref{proposition-equivalence}). 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 $\varphi$. \medskip\noindent The final statement follows for example from Lemma \ref{lemma-neighbourhood-equivalence}.
If $M$ is annihilated by $f^n$, then $$ M \otimes_R R' \cong M \otimes_{R/f^nR} R'/f^n R' \cong M \otimes_{R/f^nR} R/f^n R \cong M. $$ Since $M = \bigcup M[f^n]$ and since tensor products commute with direct limits (Algebra, Lemma \ref{algebra-lemma-tensor-products-commute-with-limits}), we obtain the desired isomorphism. The last statement is a special case of the first statement by Lemma \ref{lemma-same-quotients}.
If $M \neq 0$ but $M \otimes_R R_f = 0$, then $M$ is $f$-power torsion. By Lemma \ref{lemma-torsion-completion} we find that $M \otimes_R R' \cong M \neq 0$. The last statement is a special case of the first statement by Lemma \ref{lemma-same-quotients}.
Recall that $\Spec(R) = V(f) \amalg D(f)$ where $V(f) = \Spec(R/fR)$ and $D(f) = \Spec(R_f)$, see Algebra, Section \ref{algebra-section-spectrum-ring} and especially Lemmas \ref{algebra-lemma-spec-closed} and \ref{algebra-lemma-standard-open}. Thus the lemma follows as the map $R \to R/fR$ factors through $R'$. The last statement is a special case of the first statement by Lemma \ref{lemma-same-quotients}.
The `only if' is clear, so we assume that $M \otimes_R (R' \oplus R_f)$ is finitely generated. In this case, by writing each generator as a sum of simple tensors, $M \otimes_R (R' \oplus R_f)$ admits a finite generating set consisting of elements of $M$. That is, there exists a morphism from a finite free $R$-module to $M$ whose cokernel is killed by tensoring with $R' \oplus R_f$; we may thus deduce $M$ is finite generated by applying Lemma \ref{lemma-BL-faithful} to this cokernel. The last statement is a special case of the first statement by Lemma \ref{lemma-same-quotients}.
Let $x \in R'_f$. Write $x = x'/f^n$ with $x' \in R'$. Write $x' = x'' + f^n y$ with $x'' \in R$ and $y \in R'$. Then we see that $(y, -x''/f^n)$ maps to $x$. Thus (1) holds. \medskip\noindent Part (2) follows from the fact that $\Ker(R \to R_f) = R[f^\infty]$. \medskip\noindent If the sequence is exact in the middle, then elements of the form $(x, 0)$ with $x \in R'[f^\infty]$ are in the image of the first arrow. This implies that $R[f^\infty] \to R'[f^\infty]$ is surjective. Conversely, assume that $R[f^\infty] \to R'[f^\infty]$ is surjective. Let $(x, y)$ be an element in the middle which maps to zero on the right. Write $y = y'/f^n$ for some $y' \in R$. Then we see that $f^n x - y'$ is annihilated by some power of $f$ in $R'$. By assumption we can write $f^nx - y' = z$ for some $z \in R[f^\infty]$. Then $y = y''/f^n$ where $y'' = y' + z$ is in the kernel of $R \to R/f^nR$. Hence we see that $y$ can be represented as $y'''/1$ for some $y''' \in R$. Then $x - y'''$ is in $R'[f^\infty]$. Thus $x - y''' = z' \in R[f^\infty]$. Then $(x, y'''/1) = (y''' + z', (y''' + z')/1)$ as desired. \medskip\noindent The last statement of the lemma is a special case of the penultimate statement by Lemma \ref{lemma-same-quotients}.
We will use the results of Lemma \ref{lemma-same-f-torsion} without further mention. The functor $M \otimes_R -$ is right exact (Algebra, Lemma \ref{algebra-lemma-tensor-product-exact}) hence we get (1). \medskip\noindent The kernel of $M \to M \otimes_R R_f = M_f$ is $M[f^\infty]$. Thus (2) follows. \medskip\noindent If the sequence is exact in the middle, then elements of the form $(x, 0)$ with $x \in (M \otimes_R R')[f^\infty]$ are in the image of the first arrow. This implies that $M[f^\infty] \to (M \otimes_R R')[f^\infty]$ is surjective. Conversely, assume that $M[f^\infty] \to (M \otimes_R R')[f^\infty]$ is surjective. Let $(x, y)$ be an element in the middle which maps to zero on the right. Write $y = y'/f^n$ for some $y' \in M$. Then we see that $f^n x - y'$ is annihilated by some power of $f$ in $M \otimes_R R'$. By assumption we can write $f^nx - y' = z$ for some $z \in M[f^\infty]$. Then $y = y''/f^n$ where $y'' = y' + z$ is in the kernel of $M \to M/f^nM$. Hence we see that $y$ can be represented as $y'''/1$ for some $y''' \in M$. Then $x - y'''$ is in $(M \otimes_R R')[f^\infty]$. Thus $x - y''' = z' \in M[f^\infty]$. Then $(x, y'''/1) = (y''' + z', (y''' + z')/1)$ as desired. \medskip\noindent If $(R \to R', f)$ is a glueing pair, then (\ref{equation-BL-cech-mod-re}) is exact in the middle for any $M$ by Algebra, Lemma \ref{algebra-lemma-tensor-product-exact}. This gives the penultimate statement of the lemma. The final statement of the lemma follows from this and the fact that $(R, f)$ is a glueing pair if and only if $(R \to R^\wedge, f)$ is a glueing pair.
From the exact sequence $0 \to R[f^n] \to R \to f^n R \to 0$ we see that it suffices to check that $R[f^n] \otimes_R R' \to R'$ is injective. By Lemma \ref{lemma-torsion-completion} we have $R[f^n] \otimes_R R' = R[f^n]$ and by Lemma \ref{lemma-same-f-torsion} we see that $R[f^n] \to R'$ is injective as $(R \to R', f)$ is a glueing pair.
Set $\overline{M} = M/M[f^\infty]$. Then $\Coker(M \to M_f) \cong \Coker(\overline{M} \to \overline{M}_f)$ hence we may and do assume that $f$ is a nonzerodivisor on $M$. In this case $M \subset M_f$ and $M_f/M = \colim M/f^nM$ where the transition maps are given by multiplication by $f$. Since formation of Tor groups commutes with colimits (Algebra, Lemma \ref{algebra-lemma-tor-commutes-filtered-colimits}) it suffices to show that $\text{Tor}^R_1(R', M/f^n M) = 0$. \medskip\noindent We first treat the case $M = R/R[f^\infty]$. By Lemma \ref{lemma-same-f-torsion} we have $M \otimes_R R' = R'/R'[f^\infty]$. From the short exact sequence $0 \to M \to M \to M/f^nM \to 0$ we obtain the exact sequence $$ \xymatrix{ \text{Tor}_1^R(R', R/R[f^\infty]) \ar[r] & \text{Tor}_1^R(R', M/f^n M) \ar[r] & R'/R'[f^\infty] \ar[dll]_{f^n} \\ R'/R'[f^\infty] \ar[r] & (R'/R'[f^\infty])/(f^n (R'/R'[f^\infty])) \ar[r] & 0 } $$ by Algebra, Lemma \ref{algebra-lemma-long-exact-sequence-tor}. Here the diagonal arrow is injective. Since the first group $\text{Tor}_1^R(R', R/R[f^\infty])$ is zero by Lemma \ref{lemma-first-tor-total}, we deduce that $\text{Tor}_1^R(R', M/f^nM) = 0$ as desired. \medskip\noindent To treat the general case, choose a surjection $F \to M$ with $F$ a free $R/R[f^\infty]$-module, and form an exact sequence $$ 0 \to N \to F/f^n F \to M/f^n M \to 0. $$ By Lemma \ref{lemma-torsion-completion} this sequence remains unchanged, and hence exact, upon tensoring with $R'$. Since $\text{Tor}^R_1(R', F/f^n F) = 0$ by the previous paragraph, we deduce  that $\text{Tor}^R_1(R', M/f^n M) = 0$ as desired.
One direction of the lemma follows from Algebra, Lemma \ref{algebra-lemma-flat-base-change}. For the other direction, assume $M \otimes_R R'$ is flat over $R'$ and $M_f$ is flat over $R_f$. Let $\tilde M$ be as in Remark \ref{remark-what-you-get-for-general-modules}. If $\tilde M$ is flat over $R$, then applying Algebra, Lemma \ref{algebra-lemma-flat-tor-zero} to the short exact sequence $0 \to \Ker(M \to \tilde M) \to M \to \tilde M \to 0$ we find that $\Ker(M \to \tilde M) \otimes_R (R' \oplus R_f)$ is zero. Hence $M = \tilde M$ by Lemma \ref{lemma-BL-faithful} and we conclude. In other words, we may replace $M$ by $\tilde M$ and assume $M$ is glueable for $(R \to R', f)$. Let $N$ be a second $R$-module. It suffices to prove that $\text{Tor}_1^R(M, N) = 0$, see Algebra, Lemma \ref{algebra-lemma-characterize-flat}. \medskip\noindent The long the exact sequence of Tors associated to the short exact sequence $0 \to R \to R' \oplus R_f \to (R')_f \to 0$ and $N$ gives an exact sequence $$ 0 \to \text{Tor}_1^R(R', N) \to \text{Tor}_1^R((R')_f, N) $$ and isomorphisms $\text{Tor}_i^R(R', N) = \text{Tor}_i^R((R')_f, N)$ for $i \geq 2$. Since $\text{Tor}_i^R((R')_f, N) = \text{Tor}_i^R(R', N)_f$ we conclude that $f$ is a nonzerodivisor on $\text{Tor}_1^R(R', N)$ and invertible on $\text{Tor}_i^R(R', N)$ for $i \geq 2$. Since $M \otimes_R R'$ is flat over $R'$ we have $$ \text{Tor}_i^R(M \otimes_R R', N) = (M \otimes_R R') \otimes_{R'} \text{Tor}_i^R(R', N) $$ by the spectral sequence of Example \ref{example-tor-change-rings}. Writing $M \otimes_R R'$ as a filtered colimit of finite free $R'$-modules (Algebra, Theorem \ref{algebra-theorem-lazard}) we conclude that $f$ is a nonzerodivisor on $\text{Tor}_1^R(M \otimes_R R', N)$ and invertible on $\text{Tor}_i^R(M \otimes_R R', N)$. Next, we consider the exact sequence $0 \to M \to M \otimes_R R' \oplus M_f \to M \otimes_R (R')_f \to 0$ coming from the fact that $M$ is glueable and the associated long exact sequence of $\text{Tor}$. The relevant part is $$ \xymatrix{ \text{Tor}_1^R(M, N) \ar[r] & \text{Tor}_1^R(M \otimes_R R', N) \ar[r] & \text{Tor}_1^R(M \otimes_R (R')_f, N) \\ & \text{Tor}_2^R(M \otimes_R R', N) \ar[r] & \text{Tor}_2^R(M \otimes_R (R')_f, N) \ar[llu] } $$ We conclude that $\text{Tor}_1^R(M, N) = 0$ by our remarks above on the action on $f$ on $\text{Tor}_i^R(M \otimes_R R', N)$.
Assume that $M \otimes_R R'$ is a finite projective module over $R'$ and that $M_f$ is a finite projective module over $R_f$. Our task is to prove that $M$ is finite projective over $R$. We will use Algebra, Lemma \ref{algebra-lemma-finite-projective} without further mention. By Lemma \ref{lemma-BL-flat} we see that $M$ is flat. By Lemma \ref{lemma-faithful-descent} we see that $M$ is finite. Choose a short exact sequence $0 \to K \to R^{\oplus n} \to M \to 0$. Since a finite projective module is of finite presentation and since the sequence remains exact after tensoring with $R'$ (by Algebra, Lemma \ref{algebra-lemma-flat-tor-zero}) and $R_f$, we conclude that $K \otimes_R R'$ and $K_f$ are finite modules. Using the lemma above we conclude that $K$ is finitely generated. Hence $M$ is finitely presented and hence finite projective.
It is clear that (2) implies (1) and that (1) is equivalent to (6). Assume (1). Let $I^\bullet$ be a K-injective complex of $A$-modules representing $K$. Condition (1) signifies that $\Hom_A(A_f, I^\bullet)$ is acyclic. Let $M^\bullet$ be a complex of $A_f$-modules representing $E$. Then $$ \Hom_{D(A)}(E, K) = \Hom_{K(A)}(M^\bullet, I^\bullet) = \Hom_{K(A_f)}(M^\bullet, \Hom_A(A_f, I^\bullet)) $$ by Algebra, Lemma \ref{algebra-lemma-adjoint-hom-restrict}. As $\Hom_A(A_f, I^\bullet)$ is a K-injective complex of $A_f$-modules by Lemma \ref{lemma-hom-K-injective} the fact that it is acyclic implies that it is homotopy equivalent to zero (Derived Categories, Lemma \ref{derived-lemma-K-injective}). Thus we get (2). \medskip\noindent A free resolution of the $A$-module $A_f$ is given by $$ 0 \to \bigoplus\nolimits_{n \in \mathbf{N}} A \to \bigoplus\nolimits_{n \in \mathbf{N}} A \to A_f \to 0 $$ where the first map sends the $(a_0, a_1, a_2, \ldots)$ to $(a_0, a_1 - fa_0, a_2 - fa_1, \ldots)$ and the second map sends $(a_0, a_1, a_2, \ldots)$ to $a_0 + a_1/f + a_2/f^2 + \ldots$. Applying $\Hom_A(-, I^\bullet)$ we get $$ 0 \to \Hom_A(A_f, I^\bullet) \to \prod I^\bullet \to \prod I^\bullet \to 0 $$ Since $\prod I^\bullet$ represents $\prod_{n \geq 0} K$ this proves the equivalence of (1) and (7). On the other hand, by construction of derived limits in Derived Categories, Section \ref{derived-section-derived-limit} the displayed exact sequence shows the object $T(K, f)$ is a representative of $R\Hom_A(A_f, K)$ in $D(A)$. Thus the equivalence of (1) and (3). \medskip\noindent There is a spectral sequence $$ E_2^{p, q} = \Ext^q_A(A_f, H^p(K)) \Rightarrow \Ext^{p + q}_A(A_f, K) $$ (details omitted). This spectral sequence degenerates at $E_2$ because $A_f$ has a length $1$ resolution by projective $A$-modules (see above) hence the $E_2$-page has only 2 nonzero rows. Thus we obtain short exact sequences $$ 0 \to \Ext^1_A(A_f, H^{p - 1}(K)) \to \Ext^p_A(A_f, K) \to \Hom_A(A_f, H^p(K)) \to 0 $$ This proves (4) and (5) are equivalent to (1).
We will use the results of Lemma \ref{lemma-hom-from-Af} without further mention. If $f \in I$, and $g \in A$, then $A_{gf}$ is an $A_f$-module hence $\Ext^n_A(A_{gf}, K) = 0$ for all $n$, hence $gf \in I$. Suppose $f, g \in I$. Then there is a short exact sequence $$ 0 \to A_{f + g} \to A_{f(f + g)} \oplus A_{g(f + g)} \to A_{gf(f + g)} \to 0 $$ because $f, g$ generate the unit ideal in $A_{f + g}$. This follows from Algebra, Lemma \ref{algebra-lemma-standard-covering} and the easy fact that the last arrow is surjective. From the long exact sequence of $\Ext$ and the vanishing of $\Ext^n_A(A_{f(f + g)}, K)$, $\Ext^n_A(A_{g(f + g)}, K)$, and $\Ext^n_A(A_{gf(f + g)}, K)$ for all $n$ we deduce the vanishing of $\Ext^n_A(A_{f + g}, K)$ for all $n$. Finally, if $f^n \in I$ for some $n > 0$, then $f \in I$ because $T(K, f) = T(K, f^n)$ or because $A_f \cong A_{f^n}$.
Proof of (1). Assume $M$ is $I$-adically complete. By Lemma \ref{lemma-hom-from-Af} it suffices to prove $\Ext^1_A(A_f, M) = 0$ and $\Hom_A(A_f, M) = 0$. Since $M = \lim M/I^nM$ and since $\Hom_A(A_f, M/I^nM) = 0$ it follows that $\Hom_A(A_f, M) = 0$. Suppose we have an extension $$ 0 \to M \to E \to A_f \to 0 $$ For $n \geq 0$ pick $e_n \in E$ mapping to $1/f^n$. Set $\delta_n = fe_{n + 1} - e_n \in M$ for $n \geq 0$. Replace $e_n$ by $$ e'_n = e_n + \delta_n + f\delta_{n + 1} + f^2 \delta_{n + 2} + \ldots $$ The infinite sum exists as $M$ is complete with respect to $I$ and $f \in I$. A simple calculation shows that $fe'_{n + 1} = e'_n$. Thus we get a splitting of the extension by mapping $1/f^n$ to $e'_n$. \medskip\noindent Proof of (2). Assume that $I = (f_1, \ldots, f_r)$ and that $T(M, f_i) = 0$ for $i = 1, \ldots, r$. By Algebra, Lemma \ref{algebra-lemma-when-surjective-to-completion} we may assume $I = (f)$ and $T(M, f) = 0$. Let $x_n \in M$ for $n \geq 0$. Consider the extension $$ 0 \to M \to E \to A_f \to 0 $$ given by $$ E = M \oplus \bigoplus Ae_n\Big/\langle x_n - fe_{n + 1} + e_n\rangle $$ mapping $e_n$ to $1/f^n$ in $A_f$ (see above). By assumption and Lemma \ref{lemma-hom-from-Af} this extension is split, hence we obtain an element $x + e_0$ which generates a copy of $A_f$ in $E$. Then $$ x + e_0 = x - x_0 + fe_1 = x - x_0 - x_1 + f^2 e_2 = \ldots $$ Since $M/f^nM = E/f^nE$ by the snake lemma, we see that $x = x_0 + fx_1 + \ldots + f^{n - 1}x_{n - 1}$ modulo $f^nM$. In other words, the map $M \to \lim M/f^nM$ is surjective as desired.
Part (2) is immediate from Lemma \ref{lemma-hom-from-Af} and the definitions. For part (1), suppose that $M \to N$ is a map of derived complete modules. Denote $K = (M \to N)$ the corresponding object of $D(A)$. Pick $f \in I$. Then $\Ext_A^n(A_f, K)$ is zero for all $n$ because $\Ext_A^n(A_f, M)$ and $\Ext_A^n(A_f, N)$ are zero for all $n$. Hence $K$ is derived complete. By (2) we see that $\Ker(M \to N)$ and $\Coker(M \to N)$ are objects of $\mathcal{C}$. Finally, suppose that $0 \to M_1 \to M_2 \to M_3 \to 0$ is a short exact sequence of $A$-modules and $M_1$, $M_3$ are derived complete. Then it follows from the long exact sequence of $\Ext$'s that $M_2$ is derived complete. Thus $\mathcal{C}$ is a weak Serre subcategory by Homology, Lemma \ref{homology-lemma-characterize-weak-serre-subcategory}.
Assume that $M/IM$ is zero. Let $I = (f_1, \ldots, f_r)$. Let $i < r$ be the largest integer such that $N = M/(f_1, \ldots, f_i)M$ is nonzero. If $i$ does not exist, then $M = 0$ which is what we want to show. Then $N$ is derived complete as a cokernel of a map between derived complete modules, see Lemma \ref{lemma-serre-subcategory}. By our choice of $i$ we have that $f_{i + 1} : N \to N$ is surjective. Hence $$ \lim (\ldots \to N \xrightarrow{f_{i + 1}} N \xrightarrow{f_{i + 1}} N) $$ is nonzero, contradicting the derived completeness of $N$.
Let $K$ be a pseudo-coherent object of $D(A)$. By definition this means $K$ is represented by a bounded above complex $K^\bullet$ of finite free $A$-modules. Since $A$ is $I$-adically complete, it is derived complete (Lemma \ref{lemma-complete-derived-complete}). It follows that $H^n(K)$ is derived complete for all $n$, by part (1) of Lemma \ref{lemma-serre-subcategory}. This in turn implies that $K$ is derived complete by part (2) of the same lemma.
Define $K^\wedge$ by the last displayed formula of the lemma. There is a map of complexes $$ (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r}) \longrightarrow A $$ which induces a map $K \to K^\wedge$. It suffices to prove that $K^\wedge$ is derived complete and that $K \to K^\wedge$ is an isomorphism if $K$ is derived complete. \medskip\noindent Let $f \in A$. By Lemma \ref{lemma-double-localize} the object $R\Hom_A(A_f, K^\wedge)$ is equal to $$ R\Hom\left((A_f \to \prod\nolimits_{i_0} A_{ff_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{ff_{i_0}f_{i_1}} \to \ldots \to A_{ff_1\ldots f_r}), K\right) $$ If $f \in I$, then $f_1, \ldots, f_r$ generate the unit ideal in $A_f$, hence the extended alternating {\v C}ech complex $$ A_f \to \prod\nolimits_{i_0} A_{ff_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{ff_{i_0}f_{i_1}} \to \ldots \to A_{ff_1\ldots f_r} $$ is zero in $D(A)$ by Lemma \ref{lemma-extended-alternating-torsion}. (In fact, if $f = f_i$ for some $i$, then this complex is homotopic to zero by Lemma \ref{lemma-extended-alternating-homotopy-zero}; this is the only case we need.) Hence $R\Hom_A(A_f, K^\wedge) = 0$ and we conclude that $K^\wedge$ is derived complete by Lemma \ref{lemma-hom-from-Af}. \medskip\noindent Conversely, if $K$ is derived complete, then $R\Hom_A(A_f, K)$ is zero for all $f = f_{i_0} \ldots f_{i_p}$, $p \geq 0$. Thus $K \to K^\wedge$ is an isomorphism in $D(A)$.
By Lemma \ref{lemma-derived-completion} we know that derived completion is given by $R\Hom_A(C, -)$ for some $C \in D(A)$. Then \begin{align*} R\Hom_A(C, R\Hom_A(K, L)) & = R\Hom_A(C \otimes_A^\mathbf{L} K, L) \\ & = R\Hom_A(K, R\Hom_A(C, L)) \end{align*} by Lemma \ref{lemma-internal-hom}. This proves the first equation. The map $K \to K^\wedge$ induces a map $$ R\Hom_A(K^\wedge, L^\wedge) \to R\Hom_A(K, L^\wedge) $$ which is an isomorphism in $D(A)$ by definition of the derived completion as the left adjoint to the inclusion functor.
If (2) holds, then $K$ is derived complete with respect to $I$ by Lemma \ref{lemma-koszul-derived-completion-complete}. Conversely, assume that $K$ is derived complete with respect to $I$. Consider the filtrations $$ K_n^\bullet \supset \sigma_{\geq -r + 1}K_n^\bullet \supset \sigma_{\geq -r + 2}K_n^\bullet \supset \ldots \supset \sigma_{\geq -1}K_n^\bullet \supset \sigma_{\geq 0}K_n^\bullet = A $$ by stupid truncations (Homology, Section \ref{homology-section-truncations}). Because the construction $R\lim(K \otimes E)$ is exact in the second variable (Lemma \ref{lemma-tensor-Rlim-exact}) we see that it suffices to show $$ R\lim \left( K \otimes_A^\mathbf{L} (\sigma_{\geq p}K_n^\bullet/ \sigma_{\geq p + 1}K_n^\bullet) \right) = 0 $$ for $p < 0$. The explicit description of the Koszul complexes above shows that $$ R\lim \left( K \otimes_A^\mathbf{L} (\sigma_{\geq p}K_n^\bullet/ \sigma_{\geq p + 1}K_n^\bullet) \right) = \bigoplus\nolimits_{i_1, \ldots, i_{-p}} T(K, f_{i_1}\ldots f_{i_{-p}}) $$ which is zero for $p < 0$ by assumption on $K$.
[First proof] The assignment $K \leadsto K'$ is a functor and $K'$ is derived complete with respect to $I$ by Lemma \ref{lemma-koszul-derived-completion-complete}. By a formal argument (omitted) we see that it suffices to show $K \to K'$ is an isomorphism if $K$ is derived complete with respect to $I$. This is Lemma \ref{lemma-characterize-derived-complete-Koszul}.
Choose generators $f_1, \ldots, f_r$ of $I$. Denote $K_n$ the Koszul complex on $f_1^n, \ldots, f_r^n$ over $A$. Recall that $K_n$ is bounded and that the cohomology modules of $K_n$ are annihilated by $f_1^n, \ldots, f_r^n$ and hence by $I^{nr}$. By Lemma \ref{lemma-derived-vanishing-mod-I} we see that $K \otimes_A^\mathbf{L} K_n = 0$. Since $K$ is derived complete by Lemma \ref{lemma-derived-completion-koszul} we have $K = R\lim K \otimes_A^\mathbf{L} K_n = 0$ as desired.
Say $I$ is generated by $f_1, \ldots, f_r$. For any $K \in D(A)$ by Lemma \ref{lemma-derived-completion-koszul} we have $K^\wedge = R\lim K \otimes_A^\mathbf{L} K_n$ where $K_n$ is the Koszul complex on $f_1^n, \ldots, f_r^n$ and hence we obtain a short exact sequence $$ 0 \to R^1\lim H^{i - 1}(K \otimes_A^\mathbf{L} K_n) \to H^i(K^\wedge) \to \lim H^i(K \otimes_A^\mathbf{L} K_n) \to 0 $$ by Lemma \ref{lemma-break-long-exact-sequence-modules}. \medskip\noindent Proof of (1). Pick a distinguished triangle $K \to L \to C \to K[1]$. Then $H^i(C) = 0$ for $i \geq 0$. Since $K_n$ is sitting in degrees $\leq 0$ we see that $H^i(C \otimes_A^\mathbf{L} K_n) = 0$ for $i \geq 0$ and that $H^{-1}(C \otimes_A^\mathbf{L} K_n) = H^{-1}(C) \otimes_A A/(f_1^n, \ldots, f_r^n)$ is a system with surjective transition maps. The displayed equation above shows that $H^i(C^\wedge) = 0$ for $i \geq 0$. Applying the distinguished triangle $K^\wedge \to L^\wedge \to C^\wedge \to K^\wedge[1]$ we get (1). \medskip\noindent Proof of (2). Pick a distinguished triangle $K \to L \to C \to K[1]$. Then $H^i(C) = 0$ for $i < 0$. Since $K_n$ is sitting in degrees $-r, \ldots, 0$ we see that $H^i(C \otimes_A^\mathbf{L} K_n) = 0$ for $i < -r$. The displayed equation above shows that $H^i(C^\wedge) = 0$ for $i < r$. Applying the distinguished triangle $K^\wedge \to L^\wedge \to C^\wedge \to K^\wedge[1]$ we get (2).