Proof. We need to show that the left adjoint functor $\mathcal {C} \to {\text {Cond}}(\mathcal {C})$ is fully faithful. We take a strong limit cardinal $\kappa _1$ such that $\kappa _1>\aleph _1$ , and a regular cardinal $\kappa _2$ such that $\kappa _2>\kappa _1$ . Then there are natural functors

$$ \begin{align*}\mathcal{C}={\text{Cond}}(\mathcal{C})_{\aleph_0}\to {\text{Cond}}(\mathcal{C})={\text{Cond}}(\mathcal{C})_{\aleph_1}\to {\text{Cond}}(\mathcal{C})_{\kappa_1}\to {\text{Cond}}(\mathcal{C})_{\kappa_2},\end{align*} $$

where ${\text {Cond}}(\mathcal {C})_{\kappa }$ is defined in [Reference Mann12, Definitions 2.1.1 and 2.1.3]. Then the functor ${\text {Cond}}(\mathcal {C})_{\aleph _0}\to {\text {Cond}}(\mathcal {C})_{\kappa _1}$ is fully faithful by [Reference Aoki3, Proposition 2.5]. Moreover, by [Reference Mann12, Proposition 2.1.8], ${\text {Cond}}(\mathcal {C})_{\aleph _1}\to {\text {Cond}}(\mathcal {C})_{\kappa _2}$ is also fully faithful. Therefore, $\mathcal {C} \to {\text {Cond}}(\mathcal {C})$ is also fully faithful.

Proof. It is clear that (2) implies (1), and (3) implies (2). That (1) implies (3) can be proved similarly to [Reference Andreychev1, Proposition 5.35].

Proof. Let $M\in \mathcal {D}(A)$ be a Banach object. We will prove that $\tau ^{0\geq }M$ and $\tau ^{1\leq }M$ are Banach objects. By [Reference Mann12, Lemma 2.12.4], $\tau ^{0\geq }M$ and $\tau ^{1\leq }M$ are t-adically complete, so it is enough to show that $(\tau ^{0\geq }M)/^{\mathbb {L}}t$ and $(\tau ^{1\leq }M)/^{\mathbb {L}}t$ are discrete. We have a fiber sequence in ${\text {Cond}}(\mathcal {D}({\text {Ab}}))$

$$ \begin{align*}(\tau^{0\geq}M)/^{\mathbb{L}}t \to M/^{\mathbb{L}}t \to (\tau^{1\leq}M)/^{\mathbb{L}}t.\end{align*} $$

Since $\tau ^{-1\geq }\left ((\tau ^{0\geq }M)/^{\mathbb {L}}t\right ) \simeq \tau ^{-1\geq }(M/^{\mathbb {L}}t)$ and $\tau ^{1\leq }\left ((\tau ^{1\leq }M)/^{\mathbb {L}}t\right ) \simeq \tau ^{1\leq }(M/^{\mathbb {L}}t)$ are discrete, it is enough to show that $M_0:= H^0\left ((\tau ^{0\geq }M)/^{\mathbb {L}}t\right )$ and $M_1:= H^0\left ((\tau ^{1\leq }M)/^{\mathbb {L}}t\right )$ are discrete. Let $M_{0,{\text {disc}}}$ and $M_{1,{\text {disc}}}$ denote the discretizations of $M_0$ and $M_1$ . Then we have the following commutative diagram of static condensed $\mathbb {Z}$ -modules:

where the horizontal sequences are exact. Therefore, it is enough to show that for any static condensed abelian group $N\in ({\text {Cond}}(\mathcal {D}({\text {Ab}})))^{\heartsuit }$ , the natural morphism $N_{{\text {disc}}} \to N$ is injective. Let S be any light profinite set. We write S as a countable cofiltered limit $ S=\varprojlim\limits _{n\in \mathbb {N}} S_{n}$ , where $S_{n}$ is a finite set and the natural maps $S\to S_{n}$ are surjective. Then the map $N_{{\text {disc}}}(S) =\varinjlim\limits _{n} N(S_n) \hookrightarrow N(S)$ is injective. Therefore, $N_{{\text {disc}}} \to N$ is also injective.

Proof. It is enough to show that a Banach object $M\in \mathcal {D}(A)$ is $(A,A_0)_{\square }$ -complete and nuclear as an object of $\mathcal {D}((A,A_0)_{\square })$ . Since M can be written as a limit of discrete objects in $\mathcal {D}(A)$ and every discrete object in $\mathcal {D}(A)$ is $(A,A_0)_{\square }$ -complete, M is also $(A,A_0)_{\square }$ -complete. We will prove that M is a nuclear object of $\mathcal {D}((A,A_0)_{\square })$ . Since $ M\simeq \varinjlim\limits _{n}\tau ^{n\geq }M$ and $\tau ^{n\geq }M$ is Banach by Lemma 3.8, we may assume that M is bounded above. Let S be a light profinite set. We will show that

$$ \begin{align*}\alpha \colon R{\underline{\mathrm{Hom}}}_A((A,A_0)_{\square}[S],A) \otimes_{(A,A_0)_{\square}}^{\mathbb{L}} M \to R{\underline{\mathrm{Hom}}}_A((A,A_0)_{\square}[S],M)\end{align*} $$

is an equivalence. Since both sides are t-adically complete by [Reference Mann12, Proposition 2.12.10], so it suffices to show that

$$ \begin{align*} \alpha/^{\mathbb{L}}t \colon &R{\underline{\mathrm{Hom}}}_{A/^{\mathbb{L}}t}((A/^{\mathbb{L}}t,A_0)_{\square}[S],A/^{\mathbb{L}}t) \otimes_{(A/^{\mathbb{L}}t,A_0)_{\square}}^{\mathbb{L}} M/^{\mathbb{L}}t \\ &\to R{\underline{\mathrm{Hom}}}_{A/^{\mathbb{L}}t}((A/^{\mathbb{L}}t,A_0)_{\square}[S],M/^{\mathbb{L}}t) \end{align*} $$

is an equivalence. By [Reference Mann12, Proposition 2.9.7], the discrete object $M/^{\mathbb {L}}t \in \mathcal {D}(A/^{\mathbb {L}}t)$ is nuclear as an object of $\mathcal {D}((A/^{\mathbb {L}}t, A_0)_{\square })$ , so we find that $\alpha /^{\mathbb {L}}t$ is an equivalence.

Proof. Since every nuclear object of $\mathcal {D}((A,A_0)_{\square })$ can be written as a filtered colimit of basic nuclear objects of $\mathcal {D}((A,A_0)_{\square })$ , we may assume that M is a basic nuclear object $\varinjlim\limits _{n} P_n$ of $\mathcal {D}((A,A_0)_{\square })$ , where each $P_n$ is compact as an object of $\mathcal {D}((A,A_0)_{\square })$ and each transition morphism $P_n\to P_{n+1}$ is trace-class. From the definition of trace-class maps and $P_n^{\vee } \simeq \widehat {P_n}^{\vee }$ , where $\widehat {P_n}$ is the t-adic completion of $P_n$ and $\widehat {P_n}^{\vee }$ is the dual $R{\underline {\mathrm{Hom}}}_A(\widehat {P_n},A)$ of $\widehat {P_n}$ , the transition morphisms $P_n \to P_{n+1}$ factor through $\widehat {P_n}$ , so we get that $M \simeq \varinjlim\limits _{n} \widehat {P_n}$ . By construction, $\widehat {P_n} \to \widehat {P_{n+1}}$ is a trace-class map. It is enough to show that $\widehat {P_n} \to \widehat {P_{n+1}}$ factors through a Banach object of $\mathcal {D}(A)$ . We denote $\widehat {P_n}\to \widehat {P_{n+1}}$ by $f\colon P\to Q$ . We define a Banach object $Q_b=R\varprojlim\limits _{n} (Q/^{\mathbb {L}}t^n)_{{\text {disc}}}\in \mathcal {D}(A)$ , where $(Q/^{\mathbb {L}}t^n)_{{\text {disc}}}$ is the discretization of $Q/^{\mathbb {L}}t^n$ . We will prove that f factors through $Q_b$ . Since f is trace-class, f corresponds to a morphism $A\to P^{\vee } \otimes _{(A,A_0)_{\square }}^{\mathbb {L}} Q$ . It is enough to show that

(3.1) $$ \begin{align} R{\mathrm{Hom}}_A (A,P^{\vee}\otimes_{(A,A_0)_{\square}}^{\mathbb{L}} Q_b) \to R{\mathrm{Hom}}_A (A,P^{\vee}\otimes_{(A,A_0)_{\square}}^{\mathbb{L}} Q) \end{align} $$

is an equivalence. We may assume that $P=\widehat {(A,A_0)_{\square }[S]}$ for some light profinite set S. Since both sides of (3.1) are t-adically complete by [Reference Mann12, Proposition 2.12.10], it suffices to show that

(3.2) $$ \begin{align} R{\mathrm{Hom}}_{\mathbb{Z}} (\mathbb{Z}, (P/^{\mathbb{L}}t)^{\vee}\otimes_{(A/^{\mathbb{L}}t,A_0)_{\square}}^{\mathbb{L}} (Q/^{\mathbb{L}}t)_{{\text{disc}}}) \to R{\mathrm{Hom}}_{\mathbb{Z}} (\mathbb{Z},(P/^{\mathbb{L}}t)^{\vee}\otimes_{(A/^{\mathbb{L}}t,A_0)_{\square}}^{\mathbb{L}} Q/^{\mathbb{L}}t) \end{align} $$

is an equivalence. We take a countable set I such that there exists an isomorphism $\mathbb {Z}_{\square }[S]\cong \prod\limits _I \mathbb {Z}$ . Then $(P/^{\mathbb {L}}t)^{\vee }$ is equivalent to $R{\underline {\mathrm{Hom}}}_{\mathbb {Z}}(\prod\limits _I \mathbb {Z} , A/^{\mathbb {L}}t)\simeq \bigoplus\limits _I A/^{\mathbb {L}}t$ , so the morphism (3.2) is equivalent to

$$ \begin{align*} R{\mathrm{Hom}}_{\mathbb{Z}} (\mathbb{Z}, \bigoplus\limits_I (Q/^{\mathbb{L}}t)_{{\text{disc}}}) \to R{\mathrm{Hom}}_{\mathbb{Z}} (\mathbb{Z}, \bigoplus\limits_I Q/^{\mathbb{L}}t), \end{align*} $$

which is an equivalence.

Proof. Let S be a light profinite set. First, we assume that M is nuclear as an object of $\mathcal {D}(\mathcal {B})$ . Then we have

$$ \begin{align*} &\phantom{{}\simeq {}}R{\underline{\mathrm{Hom}}}_{\mathcal{A}} (\mathcal{A}[S], M) \\ &\simeq R{\underline{\mathrm{Hom}}}_{\mathcal{B}} (\mathcal{B}[S], M) \\ &\simeq R{\underline{\mathrm{Hom}}}_{\mathcal{B}} (\mathcal{B}[S], \underline{\mathcal{B}})\otimes_{\mathcal{B}}^{\mathbb{L}} M \\ &\simeq R{\underline{\mathrm{Hom}}}_{\mathcal{A}} (\mathcal{A}[S], \underline{\mathcal{B}})\otimes_{\mathcal{B}}^{\mathbb{L}} M \\ &\simeq \left(R{\underline{\mathrm{Hom}}}_{\mathcal{A}} (\mathcal{A}[S], \underline{\mathcal{A}})\otimes_{\mathcal{A}}^{\mathbb{L}} \underline{\mathcal{B}} \right)\otimes_{\mathcal{B}}^{\mathbb{L}} M \\ &\simeq R{\underline{\mathrm{Hom}}}_{\mathcal{A}} (\mathcal{A}[S], \underline{\mathcal{A}})\otimes_{\mathcal{A}}^{\mathbb{L}} M, \end{align*} $$

where the second equivalence follows from nuclearity of M as an object of $\mathcal {D}(\mathcal {B})$ and the fourth equivalence follows from nuclearity of $\underline {\mathcal {B}}$ as an object of $\mathcal {D}(\mathcal {A})$ , so M is nuclear as an object of $\mathcal {D}(\mathcal {A})$ .

Second, we assume that M is nuclear as an object of $\mathcal {D}(\mathcal {A})$ . Then we have

$$ \begin{align*} &\phantom{{}\simeq {}}R{\underline{\mathrm{Hom}}}_{\mathcal{B}} (\mathcal{B}[S], \underline{\mathcal{B}})\otimes_{\mathcal{B}}^{\mathbb{L}} M \\ &\simeq R{\underline{\mathrm{Hom}}}_{\mathcal{A}} (\mathcal{A}[S], \underline{\mathcal{B}})\otimes_{\mathcal{B}}^{\mathbb{L}} M \\ &\simeq \left(R{\underline{\mathrm{Hom}}}_{\mathcal{A}} (\mathcal{A}[S], \underline{\mathcal{A}})\otimes_{\mathcal{A}}^{\mathbb{L}} \underline{\mathcal{B}}\right)\otimes_{\mathcal{B}}^{\mathbb{L}} M \\ &\simeq R{\underline{\mathrm{Hom}}}_{\mathcal{A}} (\mathcal{A}[S], \underline{\mathcal{A}})\otimes_{\mathcal{A}}^{\mathbb{L}} M\\ &\simeq R{\underline{\mathrm{Hom}}}_{\mathcal{A}} (\mathcal{A}[S], M) \\ &\simeq R{\underline{\mathrm{Hom}}}_{\mathcal{B}} (\mathcal{B}[S], M), \end{align*} $$

where the second equivalence follows from nuclearity of $\underline {\mathcal {B}}$ as an object of $\mathcal {D}(\mathcal {A})$ and the fourth equivalence follows from nuclearity of M as an object of $\mathcal {D}(\mathcal {A})$ , so M is nuclear as an object of $\mathcal {D}(\mathcal {B})$ .

Proof. The “only if” direction follows from Proposition 3.10 and Lemma 3.8. We will prove the “if” direction. We may assume $(A,A_0)_{\square }=\mathbb {Z}[[t]]_{\square }$ by Corollary 3.11 and Lemma 3.12. It is enough to show that for every light profinite set S, functors $\mathbb {Z}[[t]]_{\square }[S]^{\vee } \otimes _{\mathbb {Z}[[t]]_{\square }}^{\mathbb {L}} -$ and $R{\underline {\mathrm{Hom}}}_{\mathbb {Z}[[t]]}(\mathbb {Z}[[t]]_{\square }[S],-)$ commute with Postnikov limits, where $\mathbb {Z}[[t]]_{\square }[S]^{\vee }=R{\underline {\mathrm{Hom}}}_{\mathbb {Z}[[t]]}(\mathbb {Z}[[t]]_{\square }[S],\mathbb {Z}[[t]])$ . It is clear that the latter functor commutes with Postnikov limits. We will prove that the functor $\mathbb {Z}[[t]]_{\square }[S]^{\vee } \otimes _{\mathbb {Z}[[t]]_{\square }}^{\mathbb {L}} -$ has finite Tor dimension $0$ (i.e., $\mathbb {Z}[[t]][S]^{\vee }$ is a flat $\mathbb {Z}[[t]]_{\square }$ -module). We take a countable set I such that there is an isomorphism $\mathbb {Z}_{\square }[S] \simeq \prod\limits _{I} \mathbb {Z}$ . By [Reference Rodríguez Camargo19, Corollary 3.4.3], $\prod\limits _{I} \mathbb {Z}$ is a flat $\mathbb {Z}_{\square }$ -module. Therefore, $\prod\limits _I \mathbb {Z}[[t]]=(\prod\limits _{I} \mathbb {Z})\otimes _{\mathbb {Z}_{\square }}\mathbb {Z}[[t]]$ is also a flat $\mathbb {Z}[[t]]_{\square }$ -module. We have

$$ \begin{align*}\mathbb{Z}[[t]]_{\square}[S]^{\vee} \simeq R{\underline{\mathrm{Hom}}}_{\mathbb{Z}[[t]]}(\prod\limits_I \mathbb{Z}[[t]],\mathbb{Z}[[t]]) \simeq \widehat{\bigoplus\limits_I} \mathbb{Z}[[t]],\end{align*} $$

where $\widehat {\bigoplus\limits _I} \mathbb {Z}[[t]]$ is the t-adic completion of $\bigoplus\limits _I \mathbb {Z}[[t]]$ . It can be written as a filtered colimit of $\prod\limits _I \mathbb {Z}[[t]]$ , so we get the claim.

Proof. By using the bar resolution of A over K, we get that

$$ \begin{align*}B\otimes_{(A,\mathbb{Z})_{\square}}^{\mathbb{L}} C \simeq \varinjlim\limits_{n \in \Delta} (B \otimes_{(K,\mathbb{Z})_{\square}}^{\mathbb{L}} A^{\otimes n-1} \otimes_{(K,\mathbb{Z})_{\square}}^{\mathbb{L}} C),\end{align*} $$

so $B\otimes _{(A,\mathbb {Z})_{\square }}^{\mathbb {L}} C$ is nuclear as an object of $\mathcal {D}((K,\mathbb {Z})_{\square })$ , where we note that a derived tensor product of nuclear objects is nuclear [Reference Clausen and Scholze6, Theorem 8.6]. Next, we will prove that $\pi _0(B\otimes _{(A,\mathbb {Z})_{\square }}^{\mathbb {L}} C)$ is an affinoid K-algebra. Since we have $\pi _0(B\otimes _{(A,\mathbb {Z})_{\square }}^{\mathbb {L}} C)\simeq \pi _0B\otimes _{(\pi _0A,\mathbb {Z})_{\square }} \pi _0C$ , we may assume that A, B, and C are affinoid K-algebras. We put $B=A\langle X_1,\ldots ,X_m\rangle /(f_1,\ldots ,f_n)$ . Then we have an isomorphism

$$ \begin{align*}(A\langle X_1,\ldots,X_m\rangle/(f_1,\ldots,f_n)) \otimes_{(A,\mathbb{Z})_{\square}}C \simeq (A\langle X_1,\ldots,X_m\rangle \otimes_{(A,\mathbb{Z})_{\square}}C)/(f_1,\ldots,f_n).\end{align*} $$

Therefore, it is enough to show that

$$ \begin{align*}A\langle X_1,\ldots,X_m\rangle \otimes_{(A,\mathbb{Z})_{\square}}C \simeq C\langle X_1,\ldots,X_m\rangle.\end{align*} $$

We take rings of definition $A_0\subset A$ and $C_0 \subset C$ such that the image of $A_0$ under $A\to C$ is contained in $C_0$ . Then it suffices to show that

(3.3) $$ \begin{align} A_0\langle X_1,\ldots,X_m\rangle \otimes_{(A_0,\mathbb{Z})_{\square}}C_0 \to C_0\langle X_1,\ldots,X_m\rangle \end{align} $$

is an equivalence. Since both sides are $\pi $ -adically complete by [Reference Mann12, Proposition 2.12.10], it is enough to show that (3.3) is an equivalence after taking modulo $\pi $ . The map (3.3) modulo $\pi $ is equivalent to

$$ \begin{align*}A_0/\pi[X_1,\ldots,X_m] \otimes_{(A_0/\pi,\mathbb{Z})_{\square}} C_0/\pi \to C_0/\pi[X_1,\ldots,X_m],\end{align*} $$

which is an isomorphism.

Proof. We may assume that $B=A\langle X_1,\ldots ,X_m\rangle $ or $B=A/I$ for some closed ideal $I\subset A$ . If $B=A\langle X_1,\ldots ,X_m\rangle $ , we have

$$ \begin{align*}A\langle X_1,\ldots,X_m\rangle\otimes_{(A,\mathbb{Z})_{\square}}^{\mathbb{L}} C\simeq C\langle X_1,\ldots,X_m\rangle\end{align*} $$

by the proof of Proposition 3.17, so we get the claim. Next, suppose that $B=A/I$ . Since A is a Noetherian Banach $\mathbb {Q}_p$ -algebra, $\underline {B}$ admits a resolution by finite free $\underline {A}$ -modules. Therefore, by [Reference Andreychev1, Lemma 5.10], we get

$$ \begin{align*}\underline{B}\otimes_{(A,\mathbb{Z})_{\square}}^{\mathbb{L}} \underline{C}\simeq (B(\ast)\otimes_{A(\ast)}^{\mathbb{L}}C(\ast))\otimes_{C(\ast)}^{\mathbb{L}}\underline{C}.\end{align*} $$

Since $A(\ast )\to C(\ast )$ is flat, we find that $B\otimes _{(A,\mathbb {Z})_{\square }}^{\mathbb {L}} C$ is static by [Reference Andreychev1, Theorem 5.9].

Proof. It is enough to show that for animated affinoid pairs $(A,A^+)$ , $(B,B^+)$ of weakly finite type over K, a morphism $f \colon A \to B$ of animated affinoid K-algebras defines a morphism $(A,A^+)_{\square } \to (B,B^+)_{\square }$ of analytic animated rings if and only if f defines a morphism $(A,A^+)\to (B,B^+)$ of animated affinoid pairs of weakly finite type over K. By Remark 3.2, we get

$$ \begin{align*} &\text{The morphism } f \text{ defines a morphism} (A,A^+)_{\square} \to (B,B^+)_{\square}. \\ \iff &\text{Any object M of } \mathcal{D}((B,B^+)_{\square}) \text{ is } (A,A^+)_{\square}-\text{complete.}\\ \iff &\text{Any static object } M\in \mathcal{D}((B,B^+)_{\square})^{\heartsuit} \text{ is } (\pi_0A,A^+)_{\square}-\text{complete.}\\ \iff &\text{Any object } M \text{ of } \mathcal{D}((\pi_0B,B^+)_{\square}) \text{ is } (\pi_0A,A^+)_{\square}-\text{complete.}\\ \iff &\text{The morphism } f \text{ defines a morphism } (\pi_0A,A^+)_{\square} \to (\pi_0B,B^+)_{\square}, \end{align*} $$

where we note that for an analytic animated ring $\mathcal {A}$ , an object M of $\mathcal {D}(\underline {\mathcal {A}})$ is $\mathcal {A}$ -complete if and only if $H^i(M)$ is $\mathcal {A}$ -complete for every $i\in \mathbb {Z}$ . By [Reference Andreychev1, Proposition 3.34], the last condition is equivalent to $\pi _0f(A^+) \subset B^+$ , which proves the proposition.

Proof. By [Reference Mann12, Proposition 2.3.19], we may assume $(A,A^+)_{\square }=(K,\mathbb {Z})_{\square }$ . Since B is nuclear as an object of $\mathcal {D}((K,\mathbb {Z})_{\square })$ , the map $(K,\mathbb {Z})_{\square } \to (B, \mathbb {Z})_{\square }$ is steady by [Reference Scholze20, Proposition 13.14]. The morphism $(B, \mathbb {Z})_{\square }\to (B,B^+)_{\square }$ can be obtained as a base change of $g\colon (B(\ast ), \mathbb {Z})_{\square }\to (B(\ast ),B^+)_{\square }$ , so it is enough to show that g is steady. It follows from [Reference Mann12, Proposition 2.9.7].

Proof. Since $B\otimes _{(A,\mathbb {Z})_{\square }}^{\mathbb {L}} C$ is $(B\otimes _{(A,\mathbb {Z})_{\square }}^{\mathbb {L}} C)_{\square }$ -complete, it is also $(A,A^+)_{\square }$ -, $(B,B^+)_{\square }$ -, and $(C,C^+)_{\square }$ -complete. Therefore, we get $B\otimes _{(A,\mathbb {Z})_{\square }}^{\mathbb {L}} C \simeq B\otimes _{(A,A^+)_{\square }}^{\mathbb {L}} C$ , which is $(B,B^+)_{\square }$ - and $(C,C^+)_{\square }$ -complete. It is enough to show that for an object M of $\mathcal {D}(B\otimes _{(A,\mathbb {Z})_{\square }}^{\mathbb {L}} C)$ , M is $((B,B^+)\otimes _{(A,A^+)}^{\mathbb {L}} (C,C^+))_{\square }$ -complete if and only if it is $(B,B^+)_{\square } \otimes _{(A,A^+)_{\square }}^{\mathbb {L}} (C,C^+)_{\square }$ -complete (equivalently, $(B,B^+)_{\square }$ - and $(C,C^+)_{\square }$ -complete). By the definition of the analytic animated ring structures and [Reference Andreychev1, Proposition 3.32], it suffices to show that the smallest ring of integral elements of $\pi _0(B\otimes _{(A,\mathbb {Z})_{\square }}^{\mathbb {L}} C)$ containing the image of the map $B^+\otimes C^+ \to \pi _0(B\otimes _{(A,\mathbb {Z})_{\square }}^{\mathbb {L}} C)$ is equal to the ring of integral elements of $(B,B^+)\otimes _{(A,A^+)}^{\mathbb {L}} (C,C^+)$ . It follows from the definition of $(B,B^+)\otimes _{(A,A^+)}^{\mathbb {L}} (C,C^+)$ .

Proof. Noting that the Koszul complex corresponding to the elements $f_1-gU_1, \ldots , f_n-gU_n$ in $A\langle U_1,\ldots ,U_n\rangle $ is quasi-isomorphic to $A\langle f_1/g, \ldots , f_n/g \rangle $ [Reference Kedlaya8, Exercise 1.9.26], we can prove it by the same way as in [Reference Andreychev1, Proposition 4.11].

Proof. We have a functor

$$ \begin{align*}F:= (- \otimes_{(\mathbb{Z}[T],\mathbb{Z})_{\square}}^{\mathbb{L}}(\mathbb{Z}[T,T^{-1}],\mathbb{Z})_{\square}) \otimes_{(\mathbb{Z}[U_1,\ldots,U_n],\mathbb{Z})_{\square}}^{\mathbb{L}} \mathbb{Z}[U_1,\ldots,U_n]_{\square}\end{align*} $$

from $\mathcal {D}((A,\mathbb {Z})_{\square })$ to $\mathcal {D}((A,\mathbb {Z})_{\square })$ . This functor is right t-exact, so we get

$$ \begin{align*}\pi_0(A\langle f_1/g,\ldots,f_n/g\rangle)\simeq \pi_0(F(\pi_0A))\simeq (\pi_0A)\langle f_1/g,\ldots,f_n/g\rangle,\end{align*} $$

where the last equivalence follows from Lemma 3.28. Next, we will prove that $A\langle f_1/g,\ldots ,f_n/g\rangle $ is nuclear as an object of $\mathcal {D}((K,\mathbb {Z})_{\square })$ . By Proposition 3.13 and right t-exactness of F, it is enough to show that $F(\pi _iA)$ is nuclear as an object of $\mathcal {D}((K,\mathbb {Z})_{\square })$ for every non-negative integer i. Since $\pi _iA$ is a $\pi _0A$ -module and two morphisms $(\mathbb {Z}[T],\mathbb {Z})_{\square }\to (\mathbb {Z}[T,T^{-1}],\mathbb {Z})_{\square }$ and $(\mathbb {Z}[U_1,\ldots ,U_n],\mathbb {Z})_{\square }\to \mathbb {Z}[U_1,\ldots ,U_n]_{\square }$ are steady, we get

$$ \begin{align*} & F(\pi_i A)\\ &\quad\simeq \pi_iA \otimes_{(\pi_0A,\mathbb{Z})_{\square}}^{\mathbb{L}}((\pi_0A,\mathbb{Z})_{\square} \otimes_{(\mathbb{Z}[T],\mathbb{Z})_{\square}}^{\mathbb{L}}(\mathbb{Z}[T,T^{-1}],\mathbb{Z})_{\square}) \otimes_{((\pi_0A,\mathbb{Z})_{\square} \otimes_{(\mathbb{Z}[T],\mathbb{Z})_{\square}}^{\mathbb{L}}(\mathbb{Z}[T,T^{-1}],\mathbb{Z})_{\square})} \\ &\quad\phantom{{}\simeq {}}(((\pi_0A,\mathbb{Z})_{\square} \otimes_{(\mathbb{Z}[T],\mathbb{Z})_{\square}}^{\mathbb{L}}(\mathbb{Z}[T,T^{-1}],\mathbb{Z})_{\square})\otimes_{(\mathbb{Z}[U_1,\ldots,U_n],\mathbb{Z})_{\square}}^{\mathbb{L}}\mathbb{Z}[U_1,\ldots,U_n]_{\square})\\ &\quad\simeq \pi_iA \otimes_{(\pi_0A,\mathbb{Z})_{\square}}^{\mathbb{L}} ((\pi_0A)\langle f_1/g, \ldots, f_n/g \rangle, \mathbb{Z}[f_1/g, \ldots, f_n/g])_{\square}. \end{align*} $$

By Lemma 3.12 and Proposition 3.13, $\pi _i A$ is nuclear as an object of $\mathcal {D}((\pi _0A,\mathbb {Z})_{\square })$ , so $F(\pi _i A)$ is also nuclear as an object of $\mathcal {D}(((\pi _0A)\langle f_1/g, \ldots , f_n/g \rangle , \mathbb {Z}[f_1/g, \ldots , f_n/g])_{\square })$ . Therefore, $F(\pi _i A)$ is nuclear as an object of $\mathcal {D}((K,\mathbb {Z})_{\square })$ by Corollary 3.11 and Lemma 3.12. Finally, the last claim follows from [Reference Andreychev1, Proposition 3.32].

Proof. It follows from the fact that

$$ \begin{align*}(\mathbb{Z}[T],\mathbb{Z})_{\square} \to \mathbb{Z}[T,T^{-1}],\mathbb{Z})_{\square}\end{align*} $$

and

$$ \begin{align*}(\mathbb{Z}[U_1,\ldots,U_n],\mathbb{Z})_{\square}\to \mathbb{Z}[U_1,\ldots,U_n]_{\square}\end{align*} $$

are steady localizations by [Reference Andreychev1, Lemma 3.8].

Proof. By Proposition 3.22, it is enough to show that the natural morphism

$$ \begin{align*}{\mathrm{Map}}_{(A,A^+)_{\square}/}((B,B^+)_{\square}, (C,C^+)_{\square})\to {\mathrm{Map}}_{(\pi_0A,A^+)_{\square}/}((\pi_0B,B^+)_{\square}, (\pi_0C,C^+)_{\square})\end{align*} $$

is an equivalence of anima. Since $(A,A^+)_{\square } \to (B, B^+)_{\square }$ and $(\pi _0A,A^+)_{\square } \to (\pi _0B, B^+)_{\square }$ are localizations, the above mapping anima are either empty or contractible. Therefore, it is enough to show that if ${\mathrm{Map}}_{(\pi _0A,A^+)_{\square }/}((\pi _0B,B^+)_{\square }, (\pi _0C,C^+)_{\square })\neq \emptyset $ , then ${\mathrm{Map}}_{(A,A^+)_{\square }/}((B,B^+)_{\square }, (C,C^+)_{\square })\neq \emptyset $ . We put

$$ \begin{align*}(B,B^+)=(A\langle f_1/g, \ldots, f_n/g \rangle, A^+\langle f_1/g, \ldots, f_n/g \rangle).\end{align*} $$

Since the image of g in $\pi _0C$ is invertible, we get $(\mathbb {Z}[T,T^{-1}],\mathbb {Z})_{\square } \to (C, C^+)_{\square };\; T \mapsto g$ . Moreover, the image of $f_1/g, \ldots , f_n/g$ in $\pi _0C$ lies in $C^+$ , we get $\mathbb {Z}[U_1, \ldots U_n]_{\square } \to (C, C^+)_{\square } ;\; U_i \mapsto f_i/g$ . Therefore, we get

$$ \begin{align*} (B,B^+)_{\square} &\simeq ((A,A^+)_{\square} \otimes_{(\mathbb{Z}[T],\mathbb{Z})_{\square}}^{\mathbb{L}}(\mathbb{Z}[T,T^{-1}],\mathbb{Z})_{\square}) \otimes_{(\mathbb{Z}[U_1,\ldots,U_n],\mathbb{Z})_{\square}}^{\mathbb{L}} \mathbb{Z}[U_1,\ldots,U_n]_{\square}\\ & \to (C,C^+)_{\square}, \end{align*} $$

which proves the lemma.

Proof. This functor is obviously essentially surjective. Moreover, it is fully faithful by Lemma 3.32.

Proof. It can be proved by the same way as in [Reference Andreychev1, Chapter 4].

Proof. It follows from Proposition 3.34 and [Reference Scholze20, Proposition 12.18].

Proof. It follows from [Reference Mikami15, Lemma 4.4] and Proposition 3.33.

Notation 4.1. For a t-adically complete condensed animated ring A such that $A/^{\mathbb {L}}t$ is discrete ( $t\in \pi _0(A)$ ), we denote $(A,\pi _0A)_{\square }$ by $A_{\square }$ .

Proof. We will show that $N_{B_0/A_0}^{\otimes n}$ is quasi-isomorphic to a complex of the form $0 \to M^0_0 \to M^1_0 \to M^2_0 \to 0,$ where $M^0_0$ is placed in cohomological degree $0$ and $M^0_0, M^1_0,M^2_0$ are projective objects of $\mathcal {D}((A_0)_{\square })^{\heartsuit }$ . Since $N_{B_0/A_0}^{\otimes n}$ is equivalent to $N_{B_0^{\otimes n}/A_0}$ by [Reference Mikami15, Corollary 4.8], where $B_0^{\otimes n}$ is the classical n-fold completed tensor product of $B_0$ over $A_0$ , we may assume $n=1$ . It is enough to show that for every static object M of $\mathcal {D}({A_0}_{\square })^{\heartsuit }$ , $R{\underline {\mathrm{Hom}}}_{A_0}(N_{B_0/A_0}, M) \simeq M\otimes _{{A_0}_{\square }}^{\mathbb {L}} {B_0}_{\square }$ lies in $\mathcal {D}({A_0}_{\square })^{[-2,0]}$ . Since we have $M\otimes _{{A_0}_{\square }}^{\mathbb {L}}{B_0}_{\square }\in \mathcal {D}({A_0}_{\square })^{\leq 0}$ , it suffices to show that $R{\underline {\mathrm{Hom}}}_{A_0}(N_{B_0/A_0}, M) \in \mathcal {D}({A_0}_{\square })^{\geq -2}$ . By the proof of [Reference Mikami15, Theorem 4.13], we can take a small $\pi $ -adic ring $A^{\prime }$ with a morphism $A^{\prime } \to A_0$ and a compact object $N^{\prime }\in \mathcal {D}(A^{\prime }_{\square })$ satisfying the following conditions:

1. 1. In $A^{\prime }$ , $\pi $ is a non-zero-divisor.

2. 2. There is an equivalence $N^{\prime }\otimes _{A^{\prime }_{\square }}^{\mathbb {L}} {A_0}_{\square } \simeq N_{B_0/A_0}$ .

3. 3. The $(A^{\prime }/\pi )_{\square }$ -module $N^{\prime }/^{\mathbb {L}}\pi $ has projective-amplitude in $[0,1]$ .

We will prove $R{\underline {\mathrm{Hom}}}_{A^{\prime }}(N^{\prime }, M) \in \mathcal {D}(A^{\prime }_{\square })^{\geq -2}$ for every static object M of $\mathcal {D}(A^{\prime }_{\square })^{\heartsuit }$ . Since $A^{\prime }$ is small, every compact object in $\mathcal {D}(A^{\prime }_{\square })$ is $\pi $ -adically complete [Reference Mikami15, Proposition 3.8]. We write M as a filtered colimit $\varinjlim\limits _{\lambda } M_{\lambda }$ of $\pi $ -adically complete objects of $\mathcal {D}(A^{\prime }_{\square })$ . Then $H^0(M_{\lambda })$ is also $\pi $ -adically complete [Reference Mikami15, Proposition 3.2], so we can write M as the filtered colimit $\varinjlim\limits _{\lambda } H^0(M_{\lambda })$ of static $\pi $ -adically complete objects. Therefore, we may assume that M is $\pi $ -adically complete. Then $R{\underline {\mathrm{Hom}}}_{A^{\prime }}(N^{\prime }, M)$ is $\pi $ -adically complete, and we have an equivalence

$$ \begin{align*}R{\underline{\mathrm{Hom}}}_{A^{\prime}}(N^{\prime}, M)/^{\mathbb{L}} \pi \simeq R{\underline{\mathrm{Hom}}}_{A^{\prime}/\pi}(N^{\prime}/^{\mathbb{L}}\pi, M/^{\mathbb{L}}\pi).\end{align*} $$

Since the functor $R\varprojlim\limits _{n\in \mathbb {N}}$ is left t-exact, it is enough to show that

$$ \begin{align*}R{\underline{\mathrm{Hom}}}_{A^{\prime}/\pi}(N^{\prime}/^{\mathbb{L}}\pi, M/^{\mathbb{L}}\pi) \in \mathcal{D}((A^{\prime}/\pi)_{\square})^{\geq -2}.\end{align*} $$

It follows from (3).

Proof. First, we assume that $\underline {\mathcal {A}}\in \mathcal {C}(N,n)$ . By the same argument in [Reference Mann12, The proof of $(i)\Rightarrow(ii)$ in Proposition 2.6.5], we can show that for $M\in \mathcal {C}(N,m)$ and composable morphisms $f_1,\ldots ,f_m$ , if $N\otimes _{\mathcal {A}}^{\mathbb {L}} f_i$ is zero for each i, then $M\otimes _{\mathcal {A}}^{\mathbb {L}}(f_1\circ \dots \circ f_m)$ is also zero. In particular, for composable morphisms $f_1,\ldots ,f_n$ , if $N\otimes _{\mathcal {A}}^{\mathbb {L}} f_i$ is zero for each i, then the composition $f_1\circ \dots \circ f_n$ is also zero. Let $L={\text {cofib}}(N\to \underline {\mathcal {A}})$ . We have $\mathcal {K}_f\simeq R{\underline {\mathrm{Hom}}}_{\mathcal {A}}(L,-).$ Then f is descendable of index $\leq n$ if and only if $\underline {\mathcal {A}}\to L^{\otimes n}$ is zero. The natural morphism $-\otimes _{\mathcal {A}}^{\mathbb {L}}\mathcal {B} \to (-\otimes _{\mathcal {A}}^{\mathbb {L}}\mathcal {B})\otimes _{\mathcal {A}}^{\mathbb {L}}\mathcal {B}$ admits a retraction. In other words, the natural morphism $R{\underline {\mathrm{Hom}}}_{\mathcal {A}}(N,-)\to R{\underline {\mathrm{Hom}}}_{\mathcal {A}}(N^{\otimes 2},-)$ admits a retraction. Therefore, the natural morphism $N^{\otimes 2}\to N$ admits a section. Since there is a fiber sequence

$$ \begin{align*}N^{\otimes 2}\to N \to N\otimes_{\mathcal{A}}^{\mathbb{L}}L,\end{align*} $$

we find that $N \to N\otimes _{\mathcal {A}}^{\mathbb {L}}L$ is zero. Similarly, for any $i\geq 0$ , $N\otimes _{\mathcal {A}}^{\mathbb {L}}L^{\otimes i} \to N\otimes _{\mathcal {A}}^{\mathbb {L}}L^{\otimes i+1}$ is also zero. Therefore, $\underline {\mathcal {A}}\to L\to \cdots \to L^{\otimes n}$ is also zero.

Next, we assume that f is descendable of index $\leq n$ . Then the morphism $\underline {\mathcal {A}}\to L^{\otimes n}$ is zero. We set $\mathcal {C}(N)=\bigcup _m\mathcal {C}(N,m)$ , which is closed under tensor products, retracts, shifts, and fibers. Let us prove $\underline {\mathcal {A}} \in \mathcal {C}(N)$ . Since $\mathcal {A}$ is a retract of ${\text {fib}}(\underline {\mathcal {A}}\to L^{\otimes n})$ , it is enough to show that ${\text {fib}}(\underline {\mathcal {A}}\to L^{\otimes n}) \in \mathcal {C}(N)$ . We proceed by induction on n. The case of $n=1$ is clear. We assume that ${\text {fib}}(\underline {\mathcal {A}}\to L^{\otimes n-1}) \in \mathcal {C}(N)$ . We have a fiber sequence

$$ \begin{align*}N\otimes_{\mathcal{A}}^{\mathbb{L}}{\text{fib}}(\underline{\mathcal{A}}\to L^{\otimes n-1})\to {\text{fib}}(\underline{\mathcal{A}}\to L^{\otimes n-1}) \to {\text{fib}}(L\to L^{\otimes n}).\end{align*} $$

By the induction hypothesis, we get $N\otimes _{\mathcal {A}}^{\mathbb {L}}{\text {fib}}(\underline {\mathcal {A}}\to L^{\otimes n-1}), {\text {fib}}(\underline {\mathcal {A}}\to L^{\otimes n-1})\in \mathcal {C}(N)$ . Therefore, we get ${\text {fib}}(L\to L^{\otimes n})\in \mathcal {C}(N)$ . Moreover, we have a fiber sequence

$$ \begin{align*}N={\text{fib}}(\underline{\mathcal{A}}\to L)\to {\text{fib}}(\underline{\mathcal{A}}\to L^{\otimes n})\to {\text{fib}}(L\to L^{\otimes n}).\end{align*} $$

Since $N,{\text {fib}}(L\to L^{\otimes n})\in \mathcal {C}(N)$ , we get ${\text {fib}}(\underline {\mathcal {A}}\to L^{\otimes n}) \in \mathcal {C}(N)$ .

Proof. We will prove $\underline {\mathcal {A}} \in \mathcal {C}(N,n+1)$ . We put $N_t:= \varinjlim\limits _{[m] \in \Delta _{s,\leq t}^{{\text {op}}}}N^{\otimes (m+1)}$ and $F_t:= {\text {cofib}}(N_t\to \underline {\mathcal {A}})$ for every non-negative integer t. By [Reference Mikami15, Lemma 4.12], we have a fiber sequence

$$ \begin{align*}N_{t} \to N_{t+1} \to N^{\otimes(t+2)}[t+1].\end{align*} $$

Since $N^{\otimes r}$ lies in $\mathcal {D}(\mathcal {A})^{\leq n}$ , we get that $\tau ^{\geq n-t+1}(N_t) \to \tau ^{\geq n-t+1}(N_{t+1})$ is an equivalence and $H^{n-t}(N_t) \to H^{n-t}(N_{t+1})$ is surjective. For $k\in \mathbb {Z}$ , we have an equivalence $\varinjlim\limits _{m\geq 0}\tau ^{\geq k}N_m \simeq \tau ^{\geq k}\underline {\mathcal {A}}$ , so we find that $\tau ^{\geq 1} N_n \to \tau ^{\geq 1}\underline {\mathcal {A}}$ is an equivalence and $H^0(N_n)\to H^0(\underline {\mathcal {A}})$ is surjective. Therefore, we get that $F_n$ lies in $\mathcal {D}(\mathcal {A})^{\leq -1}$ . We have the following fiber sequences:

Since $F_n$ lies in $\mathcal {D}(\mathcal {A})^{\leq -1}$ , $\beta \circ \alpha $ is zero. Therefore, we find that $\underline {\mathcal {A}}$ is a retract of $N_n$ . We can easily prove that $N_n$ lies in $\mathcal {C}(N,n+1)$ by using [Reference Mikami15, Lemma 4.12], which proves $\underline {\mathcal {A}} \in \mathcal {C}(N,n+1)$ .

Proof. Let $\mathcal {C}$ be the full $\infty $ -subcategory of $\mathcal {D}(\mathcal {A})$ consisting of those objects M such that $\varinjlim\limits _{[m] \in \Delta _{s}^{{\text {op}}}}(N^{\otimes (m+1)} \otimes _{\mathcal {A}}^{\mathbb {L}} M) \to M$ is an equivalence. We want to show that $\underline {\mathcal {A}} \in \mathcal {C}$ . Since $\mathcal {C}$ is stable under finite colimits, shifts, tensor products, and retracts, it is enough to show that $N\in \mathcal {C}$ by Lemma 5.5. We have an augmented cosimplicial analytic animated ring $\{\mathcal {B}^{n+1} \}_{n\geq -1}$ , where $\mathcal {B}^{n+1}$ is the $(n+1)$ -fold derived tensor product of $\mathcal {B}$ over $\mathcal {A}$ . Since $\mathcal {A} \to \mathcal {B}$ is steady, we have an equivalence of functors from $\mathcal {D}(\mathcal {A})$ to $\mathcal {D}(\mathcal {A})$

$$ \begin{align*}-\otimes_{\mathcal{A}}^{\mathbb{L}} \mathcal{B}^{n+1} \simeq R{\underline{\mathrm{Hom}}}_{\mathcal{A}}(N^{\otimes(n+1)},-).\end{align*} $$

Therefore, we can extend the augmented semisimplicial object $N^{\bullet }$ in $\mathcal {D}(\mathcal {A})$ to an augmented simplicial object $\{N^{\otimes (n+1)}\}_{n\geq -1}$ . Since an augmented simplicial object $\{N^{\otimes (n+1)}\otimes _{\mathcal {A}}^{\mathbb {L}} N\}_{n\geq -1}$ is split [Reference Lurie10, Definition 4.7.2.2], we find that

$$ \begin{align*}\varinjlim\limits_{[m] \in \Delta^{{\text{op}}}}(N^{\otimes(m+1)} \otimes_{\mathcal{A}}^{\mathbb{L}} N) \to N\end{align*} $$

is an equivalence. By [Reference Lurie9, Lemma 6.5.3.7], $\varinjlim\limits _{[m] \in \Delta _s^{{\text {op}}}}(N^{\otimes (m+1)} \otimes _{\mathcal {A}}^{\mathbb {L}} N) \to N$ is also an equivalence, so N lies in $\mathcal {C}$ .

Proof. A static object M of $\mathcal {D}(\mathcal {A})$ can be regarded as a $\pi _0\mathcal {A}$ -module and we have an equivalence $M\otimes _{\mathcal {A}}^{\mathbb {L}} \mathcal {B} \simeq M\otimes _{\pi _0\mathcal {A}}^{\mathbb {L}} \pi _0\mathcal {B}$ , so $-\otimes _{\mathcal {A}}^{\mathbb {L}} \mathcal {B} \colon \mathcal {D}(\mathcal {A}) \to \mathcal {D}(\mathcal {B})$ has finite Tor-dimension $\leq n$ . Let us prove that $-\otimes _{\mathcal {A}}^{\mathbb {L}} \mathcal {B} \colon \mathcal {D}(\mathcal {A}) \to \mathcal {D}(\mathcal {B})$ preserves all small limits. It is enough to show that it preserves all direct products. Let $\{M_{\lambda }\}_{\lambda \in \Lambda }$ be a family of objects in $\mathcal {D}(\mathcal {A})$ . We will show that $H^r((\prod\limits _{\lambda }M_{\lambda })\otimes _{\mathcal {A}}^{\mathbb {L}} \mathcal {B}) \to H^r(\prod\limits _{\lambda }(M_{\lambda }\otimes _{\mathcal {A}}^{\mathbb {L}} \mathcal {B}))$ is an isomorphism for every integer r. By taking a shift, we may assume $r=0$ . Note that the direct product functor in $\mathcal {D}(\mathbb {Z}_{\square })$ is t-exact because the functor $R\Gamma (S,-)\colon \mathcal {D}(\mathbb {Z}_{\square }) \to \mathcal {D}({\text {Ab}})$ is t-exact for any light profinite set S by [Reference Rodríguez Camargo19, Theorem 3.2.4]. Therefore, the direct product functor in $\mathcal {D}(\mathcal {A})$ is also t-exact. Moreover, $-\otimes _{\mathcal {A}}^{\mathbb {L}} \mathcal {B} \colon \mathcal {D}(\mathcal {A}) \to \mathcal {D}(\mathcal {B})$ has finite Tor-dimension, so we may assume that $\{M_{\lambda }\}_{\lambda }$ is uniformly bounded, and then we may assume that $M_{\lambda }$ is static for any $\lambda $ . Then the claim follows from the fact that $-\otimes _{\pi _0(\mathcal {A})}^{\mathbb {L}} \pi _0(\mathcal {B}) \colon \mathcal {D}(\pi _0(\mathcal {A})) \to \mathcal {D}(\pi _0(\mathcal {B}))$ preserves direct products.

Proof. We may assume that $M,N \in \mathcal {D}(\mathcal {A})^{\leq 0}$ . By induction, it is enough to show that $H^0(M) \to H^0(N)$ is an equivalence. It follows from $H^0(M) \simeq H^0(M\otimes _{\mathcal {A}}^{\mathbb {L}} \pi _0\mathcal {A})$ and $H^0(N) \simeq H^0(N\otimes _{\mathcal {A}}^{\mathbb {L}} \pi _0\mathcal {A})$

Proof. By Lemma 5.8, we may assume that A and B are static. The claim about finite Tor-dimension follows from [Reference Mikami15, Proposition 2.20] and Theorem 4.3. Since the forgetful functor $\mathcal {D}(B_{\square }) \to \mathcal {D}(A_{\square })$ is conservative and preserves all small limits, so it is enough to show that

$$ \begin{align*}-\otimes_{A_{\square}}^{\mathbb{L}} B_{\square} \colon \mathcal{D}(A_{\square}) \to \mathcal{D}(A_{\square})\end{align*} $$

preserves all small limits. This follows from the description

$$ \begin{align*}-\otimes_{A_{\square}}^{\mathbb{L}} B_{\square}\simeq R{\underline{\mathrm{Hom}}}_A(N_{B/A},-) \colon \mathcal{D}(A_{\square}) \to \mathcal{D}(A_{\square}).\\[-34pt] \end{align*} $$

Proof. By Lemma 5.10 and [Reference Lurie9, Proposition 5.5.2.7], the functor F is represented by some object $N \in \mathcal {D}(A_{\square })$ . Since F preserves filtered colimits, N is a compact object of $\mathcal {D}(A_{\square })$ and, in particular, it is bounded. Let us prove that $N\otimes _{A_{\square }}^{\mathbb {L}}\pi _0A_{\square } \simeq N_{\pi _0B/\pi _0A}$ . The object $N\otimes _{A_{\square }}^{\mathbb {L}}\pi _0A_{\square }$ represents a functor

$$ \begin{align*}G \colon \mathcal{D}(\pi_0A_{\square}) \to \mathcal{D}(A_{\square}) \xrightarrow{F} {\mathrm{Ani}}.\end{align*} $$

This functor can be written as $\mathcal {D}(\pi _0A_{\square }) \xrightarrow {-\otimes _{\pi _0A_{\square }}^{\mathbb {L}} \pi _0B_{\square }} \mathcal {D}(\pi _0B_{\square }) \xrightarrow {{\mathrm{Map}}(\pi _0B,-)} {\mathrm{Ani}}$ , and it is represented by $N_{\pi _0B/\pi _0A}\in \mathcal {D}(\pi _0A_{\square })$ , so we get $N\otimes _{A_{\square }}^{\mathbb {L}}\pi _0A_{\square } \simeq N_{\pi _0B/\pi _0A}$ .

Proof. We have an equivalence of anima

$$ \begin{align*} {\mathrm{Map}}_{\mathcal{D}(A_{\square})}(N_{B/A},N_{B/A}) &\simeq {\mathrm{Map}}_{\mathcal{D}(B_{\square})}(B,N_{B/A}\otimes_{A_{\square}}^{\mathbb{L}} B_{\square})\\ &\simeq {\mathrm{Map}}_{\mathcal{D}(A_{\square})}(A,N_{B/A}\otimes_{A_{\square}}^{\mathbb{L}} B_{\square}). \end{align*} $$

Let $f \colon A \to N_{B/A}\otimes _{A_{\square }}^{\mathbb {L}} B_{\square }$ be a morphism which corresponds to ${\text {id}} \colon N_{B/A} \to N_{B/A}$ . By [Reference Mann12, Lemma A.4.8], f induces a natural transformation of functors

$$ \begin{align*}\eta\colon R{\underline{\mathrm{Hom}}}_A(N_{B/A},-) \Rightarrow -\otimes_{A_{\square}}^{\mathbb{L}}B_{\square} \colon \mathcal{D}(A_{\square}) \to \mathcal{D}(A_{\square}).\end{align*} $$

We will prove that $\eta $ is an equivalence. Both functors preserve all small limits and small colimits by Lemma 5.10 and the fact that $N_{B/A}$ is compact, so it is enough to show that for a static object $M\in \mathcal {D}(A_{\square })^{\heartsuit }$ , $\eta _M \colon R{\underline {\mathrm{Hom}}}_A(N_{B/A},M) \to M\otimes _{A_{\square }}^{\mathbb {L}}B_{\square }$ is an equivalence. Since M is a $\pi _0A_{\square }$ -module, $\eta _M$ is equivalent to

$$ \begin{align*}R{\underline{\mathrm{Hom}}}_{\pi_0A}(N_{B/A}\otimes_{A_{\square}}^{\mathbb{L}} \pi_0A_{\square},M) \to M\otimes_{\pi_0A_{\square}}^{\mathbb{L}}\pi_0B_{\square},\end{align*} $$

which is an equivalence by the fact that $N_{B/A}\otimes _{A_{\square }}^{\mathbb {L}}\pi _0A_{\square } \simeq N_{\pi _0B/\pi _0A}$ .

Proof. First, we prove the former claim. Since we have an equivalence $N_{B/A}^{\otimes r}\simeq N_{B^r/A}$ , where $B^r$ is the r-fold derived tensor product of B over A, so we may assume $r=1$ . We assume that there exists an integer $t>1$ (resp. $t>2$ ) such that $H^t(N_{B/A})\neq 0$ . Then the natural map $N_{B/A} \to \tau ^{\geq t}N_{B/A}$ is not zero. Therefore, $H^0(\tau ^{\geq t}N_{B/A} \otimes _{A_{\square }}^{\mathbb {L}} B_{\square }) \cong H^0(R{\underline {\mathrm{Hom}}}_A(N_{B/A}, \tau ^{\geq t}N_{B/A}))$ is not zero. However, the functor $-\otimes _{A_{\square }}^{\mathbb {L}} B_{\square }$ has finite Tor-dimension $\leq 1$ (resp. $\leq 2$ ) by Lemma 5.10, so it is a contradiction.

Next, we prove the latter claim. By the first claim, we find that $\varinjlim\limits _{[m] \in \Delta _{s}^{{\text {op}}}}N_{B/A}^{\otimes (m+1)}$ is bounded above. Therefore, it is enough to show that $\varinjlim\limits _{[m] \in \Delta _{s}^{{\text {op}}}}N_{B/A}^{\otimes (m+1)}\otimes _{A_{\square }}^{\mathbb {L}} \pi _0A \to \pi _0A$ is an equivalence by Lemma 5.9. Since we have an equivalence $N_{B/A}\otimes _{A_{\square }}^{\mathbb {L}} \pi _0A \simeq N_{\pi _0B/\pi _0A}$ by Lemma 5.11, it is enough to show that

(5.1) $$ \begin{align} \varinjlim\limits_{[m] \in \Delta_{s}^{{\text{op}}}}N_{\pi_0B/\pi_0A}^{\otimes(m+1)} \to \pi_0A \end{align} $$

is an equivalence, where $N_{\pi _0B/\pi _0A}^{\otimes (m+1)}$ is the $(m+1)$ -fold derived tensor product of $N_{\pi _0B/\pi _0A}$ in $\mathcal {D}(\pi _0A_{\square })$ . When A is discrete, it follows from [Reference Mikami15, Theorem 2.15]. We assume that $\pi _0A$ is an affinoid K-algebra. By [Reference Mikami15, Theorem 4.13] and Lemma 5.5, $\pi _0A_{\square } \to \pi _0B_{\square }$ is descendable, so Lemma 5.7 shows that (5.1) is an equivalence.

Proof. It follows from Theorem 5.14, Lemma 3.38, and Corollary 3.35.