Proof. Obviously $M\mapsto I^\bullet M$ is right inverse to $\operatorname {Fil}^0(-)$. Conversely, for $\operatorname {Fil}^\bullet N\in \mathsf {DF}(A[1/I])$, from the filtered multiplication map $I^\bullet \otimes \operatorname {Fil}^\bullet N\to \operatorname {Fil}^\bullet N$ one gets maps $I^m\otimes \operatorname {Fil}^n N\to \operatorname {Fil}^{m+n}N$ for all $m,n\in \mathbb {Z}$. By their compatibility one can easily see that they are all isomorphisms, so $\operatorname {Fil}^\bullet N=I^\bullet \operatorname {Fil}^0N$.

Proof. If $1$ is compact and $X\in \mathcal {C}$ is dualizable with dual $X^\vee$, then since

\begin{align*} \operatorname{Hom}(X,-)=\operatorname{Hom}(1,\operatorname{\underline{Hom}}(X,-))=\operatorname{Hom}(1,X^\vee\otimes-) \end{align*}

commutes with filtered colimits, $X$ is compact.

If $\mathcal {C}$ is stable, then finite colimits of dualizable objects are dualizable: for finite direct sums it is obvious, and for cofibers one easily verify that $\operatorname {cofib}(X\to Y)^\vee =\operatorname {fib}(Y^\vee \to X^\vee )$. In general, retracts of dualizable objects are dualizable. Thus, if $\mathcal {C}$ is compactly generated by its dualizable objects, then compact objects, as retracts of finite colimits of generators, are dualizable.

Proof. Note that:

• – $1_{\mathsf {DF}(R)}=R$ and $\operatorname {Hom}(R,-)=\operatorname {Fil}^0(-)$ commutes with colimits, so $1_{\mathsf {DF}(R)}$ is compact;

• – the translations $R(n)$ are all dualizable with $R(n)^\vee =R(-n)$, and form a family of compact generators of $\mathsf {DF}(R)$, since the functor $\mathsf {DF}(R)\to \mathsf {Sp}^\mathbb {Z}$, $\operatorname {Fil}^\bullet N\mapsto (\operatorname {Fil}^nN)_{n\in \mathbb {Z}}$ is clearly conservative.

If the filtration of $R$ is trivial, then it is in particular complete, so every $R(n)$ is complete, with only one nonzero graded piece $\operatorname {gr}^{-n}R(n)=R$. Therefore, the compact objects in $\mathsf {DF}(R)$, as retracts of finite colimits of objects of the form $R(n)[d]$, are also complete, with only finitely many nonzero graded pieces that are all compact in $\mathsf {D}(R)$. Conversely, let $\operatorname {Fil}^\bullet M\in \mathsf {DF}(R)$ be complete with only finitely many nonzero graded pieces that are all compact in $\mathsf {D}(R)$. Then for $m\gg 0$ we have $\operatorname {Fil}^mM=0$ and $\operatorname {Fil}^{-m}M=M$. In addition, the terms $\operatorname {Fil}^mM$ are all compact in $\mathsf {D}(R)$, being a finite extension of compact objects. Now it is easy to write $\operatorname {Fil}^\bullet M$ as a finite extension of compact objects in $\mathsf {D}(R)$ with trivial filtration, which are clearly compact in $\mathsf {DF}(R)$.

Proof. If $c\in \mathcal {C}$ is dualizable, clearly so is $Fc\in \mathcal {D}$ since $F$ is symmetric monoidal. Conversely if $Fc\in \mathcal {D}$ is dualizable, let $c^\vee =\operatorname {\underline {Hom}}_\mathcal {C}(c,1_\mathcal {C})$, then since $F$ is symmetric monoidal and preserves internal Homs, $F(c^\vee )=(Fc)^\vee$. To prove that $c\in \mathcal {C}$ is dualizable, it suffices to show that the natural transformation $c^\vee \otimes _\mathcal {C}-\to \operatorname {\underline {Hom}}_\mathcal {C}(c,-)$ is an isomorphism. This can be checked after applying $F$ by conservativity, whence it becomes obvious.

Proof. By Lemma 2.19, $M\in \mathsf {DF}(\bar {A})$ is dualizable if and only if $M/p\in \mathsf {DF}(\bar {A}/p)$ is. Now the corollary follows from Corollary 2.18.

Proof. Assume $\operatorname {Fil}_\bullet N^\mathrm {HT}=\operatorname {Fil}^\bullet \bar {N}=0$; we want to prove that $N=0$. This is easy: $\operatorname {Fil}_\bullet N^\mathrm {HT}=0$ just means $\operatorname {gr}^\bullet N=0$ by definition, so the filtration is constant; then $\operatorname {Fil}^\bullet \bar {N}$ is also constant, and that it is zero implies that $N$ is zero by $I$-completeness.

Proof. It suffices to prove that the functor $N\mapsto (\operatorname {Fil}_\bullet N^\mathrm {HT},\operatorname {Fil}^\bullet \bar {N})$ satisfies the assumptions of Lemma 2.19. Note that the algebras $\bar {A}$ and $\operatorname {gr}^\bullet A$ in $\mathsf {DF}(A)$ are both dualizable, being cofibers of maps $I\to A$ and $A(1)\to A$. Hence, the proposition follows from Remark 2.20 and Proposition 2.22.

Proof. The first sentence is clear from the definition of $\mathsf {Gauge}(A,I)$, and the second sentence is clear from Remark 2.26.

Proof. Recall that the symmetric monoidal category of gauges is defined as the pullback

where the right vertical arrow is taking underlying objects and the lower horizontal arrow is base change along $\varphi$. Therefore, $M\in \mathsf {Gauge}(A,I)$ is dualizable if and only if its two images in $\operatorname {Fil}^\bullet M^{(1)}\in \mathsf {DF}(A)$ and $M\in \mathsf {D}(A)$ are dualizable. Now the proposition follows from Proposition 2.24. The two perfectnesses are equivalent because of Remark 2.15.

Recall that by Proposition 2.12, the symmetric monoidal category of $F$-gauges is defined as the equalizer

\begin{align*} {F\text{-}\mathsf{Gauge}}(A,I)\to\mathsf{Gauge}(A,I)\rightrightarrows\mathsf{D}(A) \end{align*}

where the two arrows send a gauge $(M,\operatorname {Fil}^\bullet M^{(1)})\in \mathsf {Gauge}(A,I)$ to $M\in \mathsf {D}(A)$ and $\operatorname {Fil}^\bullet M^{(1)}[1/I]\in \mathsf {DF}(A[1/I])\cong \mathsf {D}(A)$, respectively. Therefore, an $F$-gauge $M$ is dualizable if and only if it is dualizable as a gauge. Since $\bar {M}\in \mathsf {D}(\bar {A})$ is the underlying object of the Hodge–Tate specialization, the dualizability of the Hodge–Tate specialization implies that of $\bar {M}$, which, in turn, implies that of $M$ by Lemma 2.19.

Proof. By Proposition 2.36 we only need to show that the Hodge–Tate specialization and the de Rham specialization of $\mathrm {R}\Gamma _{ {\mathbb {\Delta }}}(X/A)$ are both dualizable. This is clear: both are complete with the graded pieces zero outside $[0,\dim (X)]$; the nonzero graded pieces are $\mathrm {R}\Gamma (X,\Omega ^\star _{X/\bar {A}}[-\star ])$, which is dualizable by [Aut, 0A1P].

Proof. This is simply because $A$ itself is an iterated extension of $(\pi _iA)[i]$ for $i=0,1,\ldots,n$, and each $\pi _iA$ is a classical $A$-module.

Proof. There is an obvious map from the left-hand side to the right-hand side. The fiber sequence $\tau _{>0}M\to M\to \tau _{\le 0}M$ of diagrams induces fiber sequences for both sides. The assumption implies that

\begin{align*} \mathop{\operatorname{colim}}\limits_I\lim_{\Delta}\tau_{>0}M=\lim_{\Delta}\mathop{\operatorname{colim}}\limits_I\tau_{>0}M=0. \end{align*}

Therefore, one can assume that $M=\tau _{\le 0}M$ is coconnective. For $n\in \mathbb {N}$, let $M_{\le n}$ be the restriction of $M$ to $\Delta _{\le n}\times I$, where $\Delta _{\le n}$ is the full subcategory of $\Delta$ generated by the simplices of dimension $\le n$. Then for every $n\in \mathbb {N}$, we have the following.

• – For every $i\in I$ we have

  \begin{align*} \operatorname{fib}\Big(\lim_{\Delta}M_i\to\lim_{\Delta_{\le n}}M_{\le n,i}\Big)\in\mathsf{Sp}_{\le-n}; \end{align*}
  since filtered colimits respect coconnectivity, we can take $\operatorname{colim}_I$ and get
  \begin{align*} \operatorname{fib}\Big(\mathop{\operatorname{colim}}_I\lim_{\Delta}M\to \mathop{\operatorname{colim}}_I\lim_{\Delta_{\le n}}M_{\le n}\Big)\in\mathsf{Sp}_{<-n}. \end{align*}

• – In addition, by the fact that filtered colimits respect coconnectivity,

  \begin{align*} \operatorname{fib}\Big(\lim_{\Delta}\mathop{\operatorname{colim}}_IM\to\lim_{\Delta_{\le n}}\mathop{\operatorname{colim}}_IM_{\le n}\Big)\in\mathsf{Sp}_{<-n}. \end{align*}

• – Since $\Delta _{\le n}$ is finite and in a stable $\infty$-category finite limits are just shifts of finite colimits,

  \begin{align*} \mathop{\operatorname{colim}}_I\lim_{\Delta_{\le n}}M_{\le n}=\lim_{\Delta_{\le n}}\mathop{\operatorname{colim}}_IM_{\le n}. \end{align*}

Combining, we get

\begin{align*} \operatorname{fib}\Big(\mathop{\operatorname{colim}}_I\lim_{\Delta}M\to\lim_{\Delta}\mathop{\operatorname{colim}}_IM\Big)\in\mathsf{Sp}_{\le-n}, \end{align*}

but the fiber is independent of $n$, so it is zero, and the lemma follows.

Proof. Combine Proposition 3.19 and (some shift of) Lemma 3.20.

Proof. We prove the lemma in steps.

• • For $\mathcal {F}=\operatorname {Fil}_\bullet \Omega ^{\not {D}}_-$, (4) is an isomorphism. This is because $\operatorname {Fil}_{<0}\Omega ^{\not {D}}_-=0$ and $\operatorname {gr}_n\Omega ^{\not {D}}_-=\mathrm {L}\Omega _-^n[-n]$, by [Reference Bhatt and LurieBL22a, Remark 4.9.3].

• • For $\mathcal {F}=\operatorname {gr}^\bullet { {\mathbb {\Delta }}}_-\{\star \}$, (4) is an isomorphism. This is due to the fiber sequence

  \begin{align*} \operatorname{gr}^m{{\mathbb {\Delta}}}_R\{n\}\to\operatorname{Fil}_m\widehat{\Omega^{\not{D}}_R}\to\operatorname{Fil}_{m-1}\widehat{\Omega^{\not{D}}_R} \end{align*}
  in [Reference Bhatt and LurieBL22a, Remark 5.5.8]. Here the hat denotes $p$-completion.

• • For $\mathcal {F}=\operatorname {Fil}^\bullet \hat {{ {\mathbb {\Delta }}}}_-\{\star \}$, where the hat denotes Nygaard completion, (4) is an isomorphism. This follows from forming finite extensions of $\operatorname {gr}^\bullet { {\mathbb {\Delta }}}_-\{\star \}$ and then taking limits.

• • Finally, the lemma follows from [Reference Bhatt and LurieBL22a, Constructions 7.4.1 and 8.4.1, Proposition 7.4.6, Remark 8.4.4].

Proof. We first show that, for any stack $X=\operatorname{colim}_{m\in \Delta ^\mathrm {op}}\operatorname {Spec}(R_m)$ as assumed, we have $\Omega ^{\not {D}}_X=\operatorname{colim}_n\operatorname {Fil}_n\Omega ^{\not {D}}_X$ ($p$-completely). For this, take $N\in \mathbb {N}$ such that $\pi _{>N}R_0=0$ ($p$-completely). Then by flatness, $\pi _{>N}R_n=0$ ($p$-completely) for every $n\in \mathbb {N}$. By Proposition 3.19 one can see that $\mathrm {L}\Omega ^n_{R_m}[-n]\in \mathsf {Sp}_{\le N}$ ($p$-completely), and thus so is $\operatorname {Fil}_n\Omega ^{\not {D}}_{R_m}$. Now by Remark 3.4 and Lemma 3.20, we have ($p$-completely)

\begin{align*} \Omega^{\not{D}}_X&=\lim_{m\in\Delta}\Omega^{\not{D}}_{R_m}=\lim_{m\in\Delta}\mathop{\operatorname{colim}}_n\operatorname{Fil}_n\Omega^{\not{D}}_{R_m}\\ &=\mathop{\operatorname{colim}}_n\lim_{m\in\Delta}\operatorname{Fil}_n\Omega^{\not{D}}_{R_m}=\mathop{\operatorname{colim}}_n\operatorname{Fil}_n\Omega^{\not{D}}_X. \end{align*}

With this in hand, we now prove the lemma.

• $\Omega^{\not{D}}_-$ This follows from the previous paragraph and the proof of Lemma 3.23.

• $\mathcal{H}_{{\mathbb {\Delta}}}$ This is because $\mathcal {H}_{\overline {{ {\mathbb {\Delta }}}}}$ is the $p$-completion of $\Omega ^{\not {D}}_-$ (cf. Example 3.14), and $\mathrm {WCart}$ is complete along $\mathrm {WCart}^\mathrm {HT}$.

• $\operatorname{Fil}^\bullet{{\mathbb {\Delta}}}_-\{\star\}$ For ${ {\mathbb {\Delta }}}_-\{\star \}$ this follows by twisting $\mathcal {H}_{ {\mathbb {\Delta }}}$ by $\mathcal {O}_\mathrm {WCart}\{\star \}$ and taking global sections. For $\operatorname {gr}^\bullet { {\mathbb {\Delta }}}_-\{\star \}$ this follows from the proof of Lemma 3.23. Combine them and get what we want.

Proof. View stacks affine over $Y$ as graded $R$-algebras. Then $Y_n$ corresponds to $S^{(n)}=R[x_0^{\pm 1},\ldots,x_n^{\pm 1}]$ with $\deg (x_i)=1$, while absolutely $Y_n$ is the spectrum of its $0$th grade part $S^{(n)}_0$. Since $S^{(n)}=S^{(n)}_0[x_0^{\pm 1}]$, it is easy to see that for an $S^{(n)}_0$-module $M$, the pushforward along $g_n$ of $M$ is the graded module $M[x_0^{\pm 1}]$. Hence, there is a fiber sequence of graded modules

\begin{align*} (g_n)_*\mathrm{L}_{Y_n/A}=\mathrm{L}_{S^{(n)}_0/A}\otimes_{S^{(n)}_0}S^{(n)}\to \mathrm{L}_{S^{(n)}/A}\to\mathrm{L}_{S^{(n)}/S^{(n)}_0}=S^{(n)}. \end{align*}

Now consider both $(\mathrm {L}_{S^{(n)}/A})$ and $(S^{(n)})$ as cosimplicial objects; then they are functorially made from the Čech nerve of $R\to S^{(0)}=R[x_0^{\pm 1}]$; but this has a left inverse as a map of ungraded rings, so both $(\mathrm {L}_{S^{(n)}/A})$ and $S^{(n)}$ split as cosimplicial ungraded modules, and in particular their pro-objects of partial totalizations are pro-constant ungradedly; now this forces them to also be pro-constant gradedly, and hence so is the pro-object of partial totalizations of the fiber $((g_n)_*\mathrm {L}_{Y_n/A})$.

Proof. Descend the fiber sequence in the proof of Proposition 3.29 and we have

\begin{align*} \mathrm{L}_{(\operatorname{Spec}(R)/\mathbb{G}_m)/A}=\operatorname{fib}(\mathrm{L}_{R/A}\to\mathrm{L}_{R/(\operatorname{Spec}(R)/\mathbb{G}_m)}) \end{align*}

as graded $R$-modules. Now view stacks affine over $\operatorname {Spec}(R)/\mathbb {G}_m$ as graded rings over $R$. Then by definition, the stack $\operatorname {Spec}(R)$ over $\operatorname {Spec}(R)/\mathbb {G}_m$ corresponds to the ring $R[t^{\pm 1}]$ with $\deg (t)=1$. Note that here $R=\mathrm {R}\Gamma (\operatorname {Spec}(R),\mathcal {O}_{\operatorname {Spec}(R)})$ is identified with the $0$th grade part of $R[t^{\pm 1}]$, which is $\bigoplus _{n\in \mathbb {Z}}R_nt^n$, while on the other hand $\mathrm {L}_{R/(\operatorname {Spec}(R)/\mathbb {G}_m)}=\mathrm {R}\Gamma (\operatorname {Spec}(R),\mathrm {L}_{R/(\operatorname {Spec}(R)/\mathbb {G}_m)})$ is identified with that of $\mathrm {L}_{R[t^{\pm 1}]/R}=R[t^{\pm 1}]\,d t$, which is $\bigoplus _{n\in \mathbb {Z}}R_nt^{n-1}\,d t$. Therefore, the map $\mathrm {L}_{R/A}\to \mathrm {L}_{R/(\operatorname {Spec}(R)/\mathbb {G}_m)}$ sends $r\in R_n$ to $d(rt^n)=nrt^{n-1}\,d t$, which gives the identification of $\mathrm {L}_{(\operatorname {Spec}(R)/\mathbb {G}_m)/A}$ in the proposition.

Proof. Clearly, if $R$ is $\mathbb {N}$-graded, then so is $\mathrm {L}_{R/A}$, with $0$th graded part $\mathrm {L}_{R_0/A}$, and thus so is $\mathrm {L}_{(\operatorname {Spec}(R)/\mathbb {G}_m)/A}=\operatorname {fib}(\mathrm {L}_{R/A}\to R)$, with $0$th graded part $\operatorname {fib}(0\colon \mathrm {L}_{R_0/A}\to R_0)=R_0[-1]\oplus \mathrm {L}_{R_0/A}$. Note that this only depends on $R_0$, so if we replace $R$ by another $\mathbb {N}$-graded $A$-algebra with the same $R_0$, the cotangent complex will be the same. This applies, in particular, to $R_0$, as a graded ring purely in the $0$th grade. Now by Proposition 3.29, one can commute the wedge power with the Čech descent and conclude that

\begin{align*} \mathrm{L}\Omega_{(\operatorname{Spec}(R)/\mathbb{G}_m)/A}^n=\mathrm{L}\Lambda^n\mathrm{L}_{(\operatorname{Spec}(R)/\mathbb{G}_m)/A}, \end{align*}

where the wedge power is over $\operatorname {Spec}(R)/\mathbb {G}_m$. Thus, the desired isomorphism follows from the fact that the wedge power of an $\mathbb {N}$-graded $R$-module $M$ is still $\mathbb {N}$-graded, whose $0$th graded part is just the wedge power of the $R_0$-module $M_0$.

Proof. Let $P=\mathbb {P}(L\oplus \mathcal {O}_S)$ be the compactification of $L$ and $\pi _P\colon P\to S$ be the projection. Let $j_\infty \colon L\to P$ and $i_\infty \colon S\to P$ be the inclusions of $L$ and $\infty$, respectively. Then there is a natural fiber sequence

\begin{align*} {i_\infty}_!i_\infty^!\pi_P^*\mathcal{M}\to\pi_P^*\mathcal{M}\to{j_\infty}_*\pi^*\mathcal{M}. \end{align*}

By the étale Poincaré duality, $\pi _P^*\mathcal {M}=\pi _P^!\mathcal {M}(-1)[-2]$, so $i_\infty ^!\pi _P^*\mathcal {M}=\mathcal {M}(-1)[-2]$. Now take $\mathrm {R}\Gamma _\mathrm {\acute {e}t}(P,-)$ of the above sequence, and the lemma follows from the classical computation of the étale cohomology of the projective line.

Proof. Since the étale cohomology is insensitive to thickenings, without loss of generality we can assume that $R$ is classical. Let $U=\operatorname {Spec}(R)\setminus \operatorname {Spec}(R_0)$ and let $\pi _U\colon U\to \operatorname {Spec}(R_0)$ and $j_U\colon U\to \operatorname {Spec}(R)$ denote the obvious maps. By the exact triangle

\begin{align*} {j_U}_!\pi_U^*\mathcal{M}\to\pi^*\mathcal{M}\to i_*\mathcal{M}, \end{align*}

it suffices to prove $\mathrm {R}\Gamma _\mathrm {\acute {e}t}(\operatorname {Spec}(R),{j_U}_!\pi _U^*\mathcal {M})=0$. Let $X$ be the blowup of $\operatorname {Spec}(R_0)$ in $\operatorname {Spec}(R)$, and let $D$ denote the exceptional divisor. Then $U$ is also $X\setminus D$, and it suffices to prove $\mathrm {R}\Gamma _\mathrm {\acute {e}t}(X,{j_U}_!\pi _U^*\mathcal {M})=0$, where $j_U$ now denotes the open immersion $U\to X$. Let $\mathcal {O}(-1)$ denote the invertible sheaf on $\operatorname {Proj}(R)$ as usual. Then by the construction of blowup, $X$ is the total space of $\mathcal {O}(-1)$ and $D$ is $\operatorname {Proj}(R)$ itself, so the proposition follows from Lemma 3.34.

Proof. This follows immediately from Proposition 3.35 by descent.

Proof. Since $\operatorname {Spec}(R)/\mathbb {G}_m$ and $\operatorname {Spec}(R_0)/\mathbb {G}_m$ are stacks over $R_0$, both sides are $\mathrm {arc}_p$ sheaves with respect to $R_0$, so we can assume that $R_0$ is perfectoid. Consider the Čech nerves of $\operatorname {Spec}(R)\to \operatorname {Spec}(R)/\mathbb {G}_m$ and $\operatorname {Spec}(R_0)\to \operatorname {Spec}(R_0)/\mathbb {G}_m$, from which we know that ${ {\mathbb {\Delta }}}_{\operatorname {Spec}(R)/\mathbb {G}_m,\mathrm {perf}}=\lim _{i\in \Delta }{ {\mathbb {\Delta }}}_{\operatorname {Spec}(R)\times \mathbb {G}_m^i,\mathrm {perf}}$ and similarly for ${ {\mathbb {\Delta }}}_{\operatorname {Spec}(R_0)/\mathbb {G}_m,\mathrm {perf}}$. Therefore, the following hold.

• – We can assume that $R$ is classical.

• – By Lemma 3.20, we can pass to limit and assume that $R$ is finitely presented over $R_0$. Say $R=\pi _0(R_0[x_1,\ldots,x_a]/(f_1,\ldots,f_b))$, where $x_1,\ldots,x_a$ and $f_1,\ldots,f_b$ are homogeneous in positive grade.

• – We can replace $R$ by the non-classical $R_0[x_1,\ldots,x_a]/(f_1,\ldots,f_b)$.

Now we are in the case that $R_0$ and $R$ are homotopically bounded $p$-quasisyntomic rings: $R_0$ is perfectoid and, hence, $p$-quasisyntomic, and $R$ is a derived complete intersection over $R_0$. Therefore, by Lemma 3.24, we have ${ {\mathbb {\Delta }}}_{\operatorname {Spec}(R)/\mathbb {G}_m}={ {\mathbb {\Delta }}}_{\operatorname {Spec}(R_0)/\mathbb {G}_m}$. Reducing modulo $p$, since the Frobenius is functorially nullhomotopic in positive homotopical degrees (see the proof of [Reference Bhatt and ScholzeBS22, Lemma 8.4], or use [Reference Bhatt and ScholzeBS17, Remark 11.8], since ${ {\mathbb {\Delta }}}_-$ actually has derived rings as values), by Lemma 3.20 we have ${ {\mathbb {\Delta }}}_{\operatorname {Spec}(R)/\mathbb {G}_m,\mathrm {perf}}/p={ {\mathbb {\Delta }}}_{\operatorname {Spec}(R_0)/\mathbb {G}_m,\mathrm {perf}}/p$. Finally, by $p$-completeness of both sides we get the desired equality.

Proof. By [Reference Bhatt and MathewBM21, Corollary 6.17], both sides are $\mathrm {arc}_p$ sheaves with respect to $R_0$. Therefore, we can assume that $R_0$ is over $\mathbb {Z}_p^\mathrm {cyc}$. Since $\mathbb {Z}_p(n)\cong \mathbb {Z}_p$ on $\operatorname {Spec}(\mathbb {Q}_p^\mathrm {cyc})$, we can assume $n=0$. Again use Čech nerves to compute both sides. Since $\mathrm {R}\Gamma _\mathrm {\acute {e}t}(\operatorname {Spec}(\hat {S}[1/p]),\mathbb {Z}_p)=({ {\mathbb {\Delta }}}_{S,\mathrm {perf}}[1/I]_p^\wedge )^{\varphi =1}$ and ${ {\mathbb {\Delta }}}_{S,\mathrm {perf}}$ is coconnective for any ring $S$, the desired equality follows from Proposition 3.38 and Lemma 3.20.

Proof. Combine Lemma 3.23 and Corollaries 3.31, 3.36, and 3.40.