Proof. Let $V\subset X$ be an open subset and $s\in {\mathcal{F}}(V)$ a section. We claim that there exists $t\in {\mathcal{F}}(X)$ with $t|_{V}=s$ . Consider the set $I$ of pairs $(W,t)$ , where $W\subset X$ is an open subset containing $V$ , and $t\in {\mathcal{F}}(W)$ , such that $t|_{V}=s$ . Inclusion of open subsets induces a partial ordering on  $I$ , where we say that $(W,t)\leqslant (W^{\prime },t^{\prime })$ if $W\subset W^{\prime }$ , and $t^{\prime }|_{W}=t$ . Moreover, $I$ is inductively ordered, that is, for every totally ordered subposet $J\subset I$ , there exists a common upper bound $i\in I$ , such that we have $i\geqslant j$ for all $j\in J$ . Indeed, denoting the pair corresponding to $j\in J$ by $(W_{j},t_{j})$ , we have $W_{j}\subset W_{k}$ for $j\leqslant k$ in $J$ , and $t_{k}|_{W_{j}}=t_{j}$ . If we define $W=\bigcup _{j\in J}W_{j}$ , the fact that ${\mathcal{F}}$ is a sheaf allows us to define a section $t\in {\mathcal{F}}(W)$ with $t|_{W_{j}}=t_{j}$ . In particular, $(W,t)\in I$ is a common upper bound for the elements of  $J$ .

Zorn’s lemma implies that the poset $I$ has a maximal element $(W,t)$ . It remains to show that $W=X$ . Assume that there exists $x\in X\setminus W$ . By assumption, $x$ has an open neighbourhood  $U$ , such that ${\mathcal{F}}|_{U}$ is flasque. In particular, there exists a section $r\in {\mathcal{F}}(U)$ , such that $r|_{U\cap W}=t|_{U\cap W}$ . By virtue of the sheaf property we obtain a section $t^{\prime }\in {\mathcal{F}}(W\cup U)$ , satisfying $t^{\prime }|_{W}=t$ , which contradicts maximality of $(W,t)$ .◻

Proof. For every point $x\in X$ there exists a neighbourhood $U_{x}$ , such that $\mathsf{M}|_{U_{x}}$ is finitely generated. Since $X$ is quasi-compact, we may choose a finite subcover $X=\bigcup _{i=1}^{n}U_{i}$ , and generating sections $(s_{ij})_{j=1,\ldots n_{i}}$ . Because $\mathsf{M}$ is assumed to be flasque, we may extend each $s_{ij}$ to a global section $t_{ij}$ , and see that this finite subset of $\unicode[STIX]{x1D6E4}(X,\mathsf{M})$ generates  $\mathsf{M}$ .◻

Proof. Since flasque sheaves have no higher cohomology, we have $H^{1}(X,\mathsf{M}_{2})=0$ , and therefore the following commutative diagram has exact rows

Commutativity of the right-hand square, and the fact that $\mathsf{M}_{1}(X){\twoheadrightarrow}\mathsf{M}_{1}(U){\twoheadrightarrow}\mathsf{M}_{0}(U)$ is surjective, implies surjectivity of $\mathsf{M}_{0}(X){\twoheadrightarrow}\mathsf{M}_{0}(U)$ .◻

Proof. Assume that we have a short exact sequence of $\mathsf{A}$ -modules $\mathsf{M}_{1}{\hookrightarrow}\mathsf{M}_{2}{\twoheadrightarrow}\mathsf{M}_{3}$ , with $\mathsf{M}_{i}$ flasque for $i=1$ and $i=3$ . Since flasque sheaves do not have higher cohomology, we see that for every open subset $U\subset X$ we have a short exact sequence of abelian groups $\mathsf{M}_{1}(U){\hookrightarrow}\mathsf{M}_{2}(U){\twoheadrightarrow}\mathsf{M}_{3}(U)$ . In particular, we obtain a commutative diagram with exact rows

with the left and right vertical arrows being surjective. The snake lemma or a simple diagram chase reveal that the vertical map in the middle also has to be surjective. This proves that $\mathsf{M}_{2}$ is a flasque sheaf.◻

Proof. According to a theorem of Keller [Reference KellerKel96, Theorem 12.1] it suffices to check that every short exact sequence of $\mathsf{A}$ -modules $\mathsf{M}_{1}{\hookrightarrow}\mathsf{M}_{2}{\twoheadrightarrow}\mathsf{M}_{3}$ with $\mathsf{M}_{1}$ flasque fits into a commutative diagram with exact rows

with $\mathsf{M}_{2}^{\prime }$ and $\mathsf{M}_{3}^{\prime }$ flasque. To produce this diagram, we recall that every $\mathsf{A}$ -module $\mathsf{M}$ can be embedded into a flasque $\mathsf{A}$ -module. Indeed, the sheaf of discontinuous sections, that is, $\mathsf{M}^{\text{dc}}(U)=\prod _{x\in U}\mathsf{M}_{x}$ provides such an embedding. Let $\mathsf{M}_{2}{\hookrightarrow}\mathsf{M}_{2}^{\prime }$ be an embedding of $\mathsf{M}_{2}$ into a flasque $\mathsf{A}$ -module. Then, the quotient $\mathsf{M}_{3}^{\prime }=\mathsf{M}_{2}^{\prime }/\mathsf{M}_{1}$ is also flasque by Lemma 2.3.◻

Proof. We will apply the dual of the result in Keller [Reference KellerKel96, Theorem 12.1], by which it suffices to check that every short exact sequence of flasque $\mathsf{A}$ -modules $\mathsf{M}_{1}{\hookrightarrow}\mathsf{M}_{2}{\twoheadrightarrow}\mathsf{M}_{3}$ with $\mathsf{M}_{3}$ projective fits into a commutative diagram with exact rows

with $\mathsf{M}_{1}^{\prime }$ and $\mathsf{M}_{2}^{\prime }$ projective. Since $\mathsf{M}_{2}\in \mathsf{Mod}_{\mathsf{fl}}(\mathsf{A})$ is flasque by assumption, there exists a surjection $\mathsf{A}^{\oplus I}\rightarrow \mathsf{M}_{2}$ for some index set  $I$ . Indeed, we can take $I=\{(U,s)|U\subset X\text{ open, }s\in \mathsf{M}_{2}(U)\}$ . Choosing an extension $t_{(U,s)}\in \mathsf{M}_{2}(X)$ , satisfying $t_{(U,s)}|_{U}=s$ for every element of  $I$ , we obtain a surjective morphism of $\mathsf{A}$ -module $\mathsf{A}^{\oplus I}\rightarrow \mathsf{M}_{2}$ .

Let $\mathsf{M}_{2}^{\prime }=\mathsf{A}^{\oplus I}$ , and define $M_{1}^{\prime }$ to be the kernel of the composition $\mathsf{M}_{2}^{\prime }\rightarrow \mathsf{M}_{2}\rightarrow \mathsf{M}_{3}$ . Since $\mathsf{M}_{3}$ is a direct summand of a free $\mathsf{A}$ -module, and $\mathsf{M}_{2}^{\prime }$ is flasque, there exists a splitting to this surjection. Therefore, $\mathsf{M}_{1}^{\prime }$ belongs to $\mathsf{P}(\mathsf{A})$ , since it is a direct summand of  $\mathsf{M}_{2}^{\prime }$ . This concludes the proof.◻

Proof. We have a short exact sequence $\ker f{\hookrightarrow}V_{1}{\twoheadrightarrow}\operatorname{im}f$ , since the first two sheaves are flasque, so is the third (Lemma 2.3). The same argument applies to the short exact sequence $\operatorname{im}f{\hookrightarrow}V_{2}{\twoheadrightarrow}\operatorname{coker}f$ , and implies that $\operatorname{coker}f$ is flasque.◻

Proof. Since every projective quasi-coherent $\mathsf{A}$ -module is a direct summand of a free $\mathsf{A}$ -module, there exist quasi-coherent $\mathsf{A}$ -modules $W_{1}$ and $W_{2}$ , such that $V_{i}\oplus W_{i}$ are free $\mathsf{A}$ -modules for $i=1,2$ . The induced map

$$\begin{eqnarray}f\oplus \operatorname{id}:V_{1}\oplus (W_{1}\oplus W_{2}\oplus V_{1}\oplus V_{2})^{\oplus \mathbb{N}}\rightarrow V_{2}\oplus (W_{1}\oplus W_{2}\oplus V_{1}\oplus V_{2})^{\oplus \mathbb{N}}\end{eqnarray}$$

has the same kernel $\ker f\simeq \ker (f\oplus \operatorname{id})$ . However, the Eilenberg swindle

$$\begin{eqnarray}(W_{1}\oplus W_{2}\oplus V_{1}\oplus V_{2})^{\oplus \mathbb{N}}\simeq (W_{1})^{i}\oplus (W_{2})^{j}\oplus (W_{1}\oplus W_{2}\oplus V_{1}\oplus V_{2})^{\oplus \mathbb{N}}\end{eqnarray}$$

allows us to see that the two sides are, in fact, free $\mathsf{A}$ -modules. Therefore, the defining property of lâche sheaf of algebras implies that $\ker f$ is flasque. Lemma 2.11 yields that $\operatorname{im}f$ and $\operatorname{coker}f$ are flasque sheaves.◻

Proof. We have seen in Lemma 2.1 that a sheaf is flasque if and only if it is locally flasque. Therefore, we may assume $M\in \mathsf{D}(\mathsf{P}(\mathsf{A}))$ . Let us choose an explicit presentation by a complex $(V^{\bullet },d)$ , where each $V^{i}$ is a projective $\mathsf{A}$ -module. We have ${\mathcal{H}}^{i}(M)\simeq (\ker d^{i})/(\operatorname{im}d^{i-1})$ . By Lemma 2.12, $\ker d^{i}$ and $\operatorname{im}d^{i-1}$ are flasque. By Lemma 2.3, the quotient ${\mathcal{H}}^{i}(M)$ is flasque.◻

Proof. For an open subset $U\subset X$ we have a map $\unicode[STIX]{x1D719}:\mathsf{A}(U)\rightarrow \mathsf{A}(X)$ which sends $s\in \mathsf{A}(U)$ to the unique section $\widehat{s}\in \mathsf{A}(X)$ , such that $\widehat{s}|_{U}=s$ , and $\widehat{s}|_{X\setminus U}=0$ . This definition makes sense because $\mathsf{A}$ is a sheaf, and $X=U\cup X\setminus U$ a disjoint open covering. Since this map is $\mathsf{A}(X)$ -linear, we have shown that $\mathsf{A}$ is very flasque.◻

Proof. Let $f:\mathsf{A}_{V}^{\oplus J}\rightarrow \mathsf{A}_{V}^{\oplus I}$ be an $\mathsf{A}_{V}$ -linear map, where $V\subset X$ is open. We have to show that $K=\ker f$ is a flasque sheaf on  $V$ . For $U\subset V$ open we have a commutative diagram

with exact rows, because taking global sections is a left exact functor. However, $\mathsf{A}(V)$ -linearity of the section $r_{V}\circ \unicode[STIX]{x1D719}_{U}:\mathsf{A}(U)\rightarrow \mathsf{A}(V)$ implies that we have a commutative diagram

where the dashed arrow is provided by the universal property of kernels. The dashed arrow is therefore right-inverse to the restriction map $K(V)\rightarrow K(U)$ , and we conclude that $K=\ker f$ is flasque.◻

Proof. Lemma 1.14 asserts that $\mathsf{A}_{X,T}({\mathcal{F}})$ is very flasque. According to Lemma 2.16 this implies that $\mathsf{A}_{X,T}({\mathcal{F}})$ is also lâche.◻

Proof. By definition, $|\mathsf{QC}(\mathsf{A}_{X}^{\bullet })|$ is the limit of a co-simplicial diagram of symmetric monoidal stable $\infty$ -categories. Therefore, it is stable (see [Reference LurieLur, Proposition 1.1.4.6]) and symmetric monoidal itself.

The $\infty$ -categories $\mathsf{QC}(\mathsf{A}_{X}^{k})=\mathsf{Ind}\mathsf{Perf}(\mathsf{A}_{X}^{k})$ are presentable for each $k\in \mathbf{N}$ (see [Reference LurieLur07, Theorem 5.5.1.1(4)]). Since a limit of presentable $\infty$ -categories is again presentable [Reference LurieLur07, Proposition 5.5.3.13], we see that this is also the case for $|\mathsf{QC}(\mathsf{A}_{X}^{\bullet })|$ . Presentable $\infty$ -categories possess small colimits by [Reference LurieLur07, Definition 5.5.0.1].◻

Proof. Let $(N_{i}^{\bullet })_{i\in I}$ be an arbitrary small family of objects in $\mathsf{QC}(\mathsf{A}_{X}^{\bullet })$ , and $f:M^{\bullet }\rightarrow \bigoplus _{i\in I}N_{i}^{\bullet }$ an arbitrary morphism. It suffices to show that there exists a finite subset $J\subset I$ , such that we have a factorisation

(4)

To produce this factorisation, we will show that the space of commutative diagrams as drawn above is either empty or contractible. Since $\mathsf{Perf}(\mathsf{A}_{X}^{0})\subset \mathsf{QC}(\mathsf{A}_{X}^{0})^{c}$ , by virtue of the definition $\mathsf{QC}(\mathsf{A}_{X}^{0})=\mathsf{Ind}\mathsf{Perf}(\mathsf{A}_{X}^{0})$ , we see that we have a finite subset $J\subset I$ and a factorisation $M^{0}\rightarrow \bigoplus _{i\in J}N_{i}^{0}\rightarrow \bigoplus _{i\in I}N_{i}^{0}$ . Since the modules $M^{\bullet }$ and $(N_{i}^{\bullet })_{i\in I}$ are cartesian, that is, $M^{k}\simeq M^{0}\otimes _{\mathsf{A}_{X}^{0}}\mathsf{A}_{X}^{k}$ , we see that we have a factorisation $M^{k}\rightarrow \bigoplus _{i\in J}N_{i}^{k}\rightarrow \bigoplus _{i\in I}N_{i}^{k}$ for all co-simplicial levels $k\geqslant 0$ .

Consider the cartesian diagram of co-simplicial spaces

where $\mathsf{pt}^{\bullet }$ is the constant diagram, consisting levelwise of a single point, and the map $\mathsf{pt}^{\bullet }\rightarrow \mathsf{Map}(M,\bigoplus _{i\in I}N_{i}^{\bullet })$ corresponds to the element $f\in |\mathsf{Map}(M,\bigoplus _{i\in I}N_{i}^{\bullet })|$ (where we used Lemma A.2). We know that the fibre $X^{\bullet }\rightarrow \mathsf{pt}^{\bullet }$ is contractible: as we have remarked above, the space of factorisations through $\bigoplus _{i\in J}N_{i}^{\bullet }\rightarrow \bigoplus _{i\in I}N_{i}^{\bullet }$ is either empty or contractible. We have shown that in the case of $f$ , a factorisation exists levelwise. Hence, we have a fibrewise equivalence $X^{\bullet }\rightarrow \mathsf{pt}^{\bullet }$ .

Taking the limit of the co-simplicial diagram above, we obtain a cartesian diagram of spaces

which implies that there exists a unique factorisation as in (4). ◻

Proof. This follows from Lemma A.6, and the fact that $-\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{X}^{\bullet }$ preserves perfect complexes, that is, compact objects (by Lemma 3.9).◻

Proof. Since $\int _{X}$ commutes with small colimits by Lemma 3.10, we may assume without loss of generality that $M\in \mathsf{Perf}(X)$ . In Lemma 3.8 we have seen that under the equivalence $\mathsf{QC}(X)=\mathsf{Ind}\mathsf{Perf}(X)$ , the functor $\int _{X}$ equivalent to $\text{}\underline{\mathsf{Hom}}_{\mathsf{A}_{X}^{\bullet }}(-\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{X}^{\bullet },-)$ , that is, $\int M\otimes N^{\bullet }$ is equivalent to the functor $(-\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{X}^{\bullet })^{\vee }\otimes _{\mathsf{A}_{X}^{\bullet }}M\otimes _{{\mathcal{O}}_{X}}N^{\bullet }\simeq M\otimes _{{\mathcal{O}}_{X}}(-\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{X}^{\bullet })^{\vee }\otimes _{\mathsf{A}_{X}^{\bullet }}N^{\bullet }$ . The right-hand side is equivalent to $M\otimes _{{\mathcal{O}}_{X}}\int N^{\bullet }$ .◻

Proof. This is a special case of Lemma 3.11, once we will have shown that $\int \mathsf{A}_{X}^{\bullet }\simeq {\mathcal{O}}_{X}$ . By definition, we have $\int _{X}\mathsf{A}_{X}^{\bullet }\simeq \text{}\underline{\mathsf{Hom}}_{\mathsf{A}_{X}^{\bullet }}(\mathsf{A}_{X}^{\bullet },\mathsf{A}_{X}^{\bullet })\simeq |\mathsf{A}_{X}^{\bullet }|$ , where the last equivalence uses the description of Hom-spaces in limits of $\infty$ -categories of Lemma A.2. By Theorem 1.16 the sheaf $|\mathsf{A}_{X}^{\bullet }|$ is canonically equivalent to ${\mathcal{O}}_{X}$ , by means of the augmentation ${\mathcal{O}}_{X}\rightarrow \mathsf{A}_{X}^{\bullet }$ .◻

Proof. The morphism of co-simplicial rings $\mathsf{A}_{X}^{\bullet }\rightarrow \mathsf{A}_{X,|Z|}^{\bullet }$ has the property that it is levelwise the projection to a factor ring. That is, for each $n\geqslant 0$ , we have a decomposition

$$\begin{eqnarray}\mathsf{A}_{X}^{n}\simeq \mathsf{A}_{X,|Z|_{n}}^{n}\times \mathsf{A}_{X,|X|_{n}\setminus |Z|_{n}}^{n}.\end{eqnarray}$$

By induction one shows that for $M^{\bullet }\in |\mathsf{Perf}_{|Z|}(\mathsf{A}_{X}^{\bullet })|,$ we have $M^{n}\otimes _{\mathsf{A}_{X}^{n}}\mathsf{A}_{X,|X|_{n}\setminus |Z|_{n}}^{n}\simeq 0$ .

This holds for $n=0$ by virtue of the assumption that $M^{0}|_{X\setminus Z}\simeq 0$ . We assume by induction that it holds for $k\leqslant n$ , and we will show that $M^{n+1}\otimes _{\mathsf{A}_{X}^{n+1}}\mathsf{A}_{X,|X|_{n+1}\setminus |Z|_{n+1}}^{n+1}\simeq 0$ . To see this, we observe that $\mathsf{A}_{X,|X|_{n+1}\setminus |Z|_{n+1}}^{n+1}$ decomposes as a product $\mathsf{A}_{X,T_{1}}^{n+1}\times \mathsf{A}_{X,T_{2}}^{n+1}$ . The subset $T_{1}\subset |X|_{n+1}$ consists of all chains $(x_{0}\leqslant \cdots \leqslant x_{n+1})$ , with $(x_{0}\leqslant \cdots \leqslant x_{n})\in |X|_{n}\setminus |Z|_{n}$ , and $x_{n+1}\in |Z|$ ; and $T_{2}\subset |X|_{n+1}$ consists of chains $(x_{0}\leqslant \cdots \leqslant x_{n+1})$ with $x_{n+1}\in |X|\setminus |Z|$ . We have equivalences

$$\begin{eqnarray}M^{n+1}\otimes _{\mathsf{A}_{X}^{n+1}}\mathsf{A}_{X,T_{1}}^{n+1}\simeq M^{n}\otimes _{\mathsf{A}_{X}^{n}}\mathsf{A}_{X,|X|_{n}\setminus |Z|_{n}}^{n}\otimes _{\mathsf{A}_{X,|X|_{n}\setminus |Z|_{n}}^{n}}\mathsf{A}_{X,T_{1}}^{n+1}\simeq 0\end{eqnarray}$$

and

$$\begin{eqnarray}M^{n+1}\otimes _{\mathsf{A}_{X}^{n+1}}\mathsf{A}_{X,T_{2}}^{n+1}\simeq M^{0}\otimes _{\mathsf{A}_{X}^{0}}\mathsf{A}_{X,|X|_{0}\setminus |Z|_{0}}^{0}\otimes _{\mathsf{A}_{X,|X|_{0}\setminus |Z|_{0}}^{0}}\mathsf{A}_{X,T_{2}}^{n+1}\simeq 0\end{eqnarray}$$

by the induction hypothesis. This implies the vanishing assertion.

We have seen that for every $n\geqslant 0$ the functor $-\otimes _{\mathsf{A}_{X}^{n}}\mathsf{A}_{X,|Z|_{n}}^{n}$ is fully faithful, when restricted to the full subcategory of $\mathsf{Perf}(\mathsf{A}_{X}^{n})$ , consisting of objects which are equivalent to 0, after tensoring with $-\otimes _{\mathsf{A}_{X}^{n}}\mathsf{A}_{X,|X|_{n}\setminus |Z|_{n}}^{n}$ . That is, for two objects $M^{\bullet }$ , $N^{\bullet }$ in this category, we have an equivalence

$$\begin{eqnarray}\mathsf{Map}(M^{n},N^{n})\simeq \mathsf{Map}(M^{n}\otimes _{\mathsf{A}_{X}^{n}}\mathsf{A}_{X,|Z|_{n}}^{n},N^{n}\otimes _{\mathsf{A}_{X}^{n}}\mathsf{A}_{X,|Z|_{n}}^{n})\end{eqnarray}$$

for every $n\geqslant 0$ . This implies that

$$\begin{eqnarray}\underset{[n]\in \unicode[STIX]{x1D6E5}}{\mathsf{lim}}\mathsf{Map}(M^{n},N^{n})\rightarrow \underset{[n]\in \unicode[STIX]{x1D6E5}}{\mathsf{lim}}\mathsf{Map}(M^{n}\otimes _{\mathsf{A}_{X}^{n}}\mathsf{A}_{X,|Z|_{n}}^{n},N^{n}\otimes _{\mathsf{A}_{X}^{n}}\mathsf{A}_{X,|Z|_{n}}^{n})\end{eqnarray}$$

is an equivalence too. Using Lemma A.2, we identify both sides with the mapping spaces in the limit $\infty$ -categories and see that the functor

$$\begin{eqnarray}-\otimes _{\mathsf{A}_{X}^{\bullet }}\mathsf{A}_{X,|Z|}^{\bullet }:|\mathsf{Perf}_{|Z|}(\mathsf{A}_{X}^{\bullet })|{\hookrightarrow}|\mathsf{Perf}(\mathsf{A}_{X,|Z|}^{\bullet })|\end{eqnarray}$$

is fully faithful. ◻

Proof. For $M^{\bullet }$ in $|\mathsf{Perf}_{|Z|}(\mathsf{A}_{X}^{\bullet })|$ , we have $\int _{X}M^{\bullet }\in \mathsf{QC}_{Z}(X)$ . Therefore, it suffices to consider only perfect complexes in $X$ with set-theoretic support $|Z|$ . The right-hand side functor agrees with $\mathsf{Map}(-\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{X}^{\bullet },-):\mathsf{Perf}_{Z}(X)^{\mathsf{op}}\times |\mathsf{Perf}_{|Z|}(\mathsf{A}_{X}^{\bullet })|\rightarrow \infty \text{-}\mathsf{Gpd}$ . By Lemma 3.14 it is equivalent to $\mathsf{Map}(-\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{X,|Z|}^{\bullet },-\otimes _{\mathsf{A}_{X}^{\bullet }}\mathsf{A}_{X,|Z|}^{\bullet })$ , hence to $\int _{X,|Z|}\circ (-\otimes _{\mathsf{A}_{X}^{\bullet }}\,\mathsf{A}_{X,|Z|}^{\bullet })$ .◻

Proof. Since $N^{0}\neq 0$ , it follows from Bhatt’s Theorem 3.16 that there is a (not necessarily closed) point $x\in X$ with $N^{0}\otimes _{\mathsf{A}_{X}^{0}}\widehat{{\mathcal{O}}}_{x}\neq 0$ . Since $N^{\bullet }$ is non-zero there exists a point $x\in X$ , such that the perfect complex of $\widehat{{\mathcal{O}}_{x}}$ -modules $N^{0}\otimes _{\mathsf{A}_{X}^{0}}\widehat{{\mathcal{O}}}_{x}$ is non-zero. The Yoneda embedding restricts to a fully faithful functor $\mathsf{Perf}(\widehat{{\mathcal{O}}}_{x}){\hookrightarrow}\mathsf{Ind}\mathsf{Perf}_{\{x\}}(X)$ (see the reformulation of [Reference EfimovEfi10] in [Reference Braunling, Groechenig and WolfsonBGW14, Propositions 5.2 and 5.4]). Therefore, there exists an object $F\in \mathsf{Perf}_{\{x\}}(X)$ , such that $\text{}\underline{\mathsf{Hom}}(F^{\vee },N^{\bullet }\otimes _{\mathsf{A}_{X}^{0}}\widehat{{\mathcal{O}}}_{x})=F\otimes _{{\mathcal{O}}_{X}}N^{0}\otimes _{\mathsf{A}_{X}^{0}}\widehat{{\mathcal{O}}}_{x}\neq 0$ . This implies $F\otimes _{{\mathcal{O}}_{X}}N^{\bullet }\neq 0$ .◻

Proof. We apply Lemma 3.20 to deduce the existence of a (not necessarily closed) point $y\in X$ , such that $N^{0}\otimes _{\mathsf{A}_{X}^{0}}\widehat{{\mathcal{O}}}_{y}\neq 0$ . We choose $F\in \mathsf{Perf}_{\{y\}}(\widehat{{\mathcal{O}}}_{y})$ , such that $N^{0}\otimes _{\mathsf{A}_{X}^{0}}\widehat{{\mathcal{O}}_{y}}\otimes _{\widehat{{\mathcal{O}}}_{y}}F\neq 0$ (existence of such a point $y$ and a complex $F$ are guaranteed by Lemma 3.20). It is clear that $N^{0}\otimes _{\mathsf{A}_{X}^{0}}\widehat{{\mathcal{O}}}_{y}\otimes _{{\mathcal{O}}_{y}}F\neq 0$ implies $N^{0}\otimes _{\mathsf{A}_{X}^{0}}\widehat{{\mathcal{O}}_{y}}\neq 0$ . This justifies making this replacement. However, note that $N^{\bullet }\otimes _{{\mathcal{O}}_{X}}F\in \mathsf{QC}(\mathsf{A}_{X}^{\bullet })$ , because $F$ might not be perfect as a complex of ${\mathcal{O}}_{X}$ -module. Nonetheless we may view $N^{0}\otimes _{{\mathcal{O}}_{X}}F$ as an object in $\mathsf{Perf}_{\{y\}}({\mathcal{O}}_{y})$ .

Let $x\in \overline{\{y\}}$ be a closed point. Observe that such an $x$ exists, because $X$ is affine (every ideal is contained in a maximal ideal). We claim that $N^{0}\otimes _{\mathsf{A}_{X}^{0}}\widehat{{\mathcal{O}}_{x}}\neq 0$ as well. Indeed, we have a map co-simplicial sheaves of algebras $f:\mathsf{A}_{X}^{\bullet }\rightarrow \mathsf{A}_{X,T_{\bullet }}^{\bullet }$ , where $T_{\bullet }\subset |X|_{\bullet }$ is the minimal simplicial subset containing the chains $(x)$ , $(y)$ , and $(x<y)$ .

By [Reference HuberHub91, Proposition 3.3.3], the ring $\mathsf{A}_{X,T_{1}}^{1}$ is isomorphic to a product $\mathsf{A}^{0}\times \mathsf{A}(x<z)$ , where the first factor corresponds to the degenerate part of the co-simplicial ring.

The maps $\mathsf{A}_{X}^{0}\rightarrow \widehat{{\mathcal{O}}}_{z}$ for $z=x,y$ factor through $\mathsf{A}_{X,T_{0}}^{0}({\mathcal{O}}_{X})$ . We have the following commutative diagram (see [Reference HuberHub91, Proposition 3.2.1] for how $\mathsf{A}(x<y)$ fits in there).

We recall that the set-theoretic support condition $N^{\bullet }\otimes _{{\mathcal{O}}_{X}}F\in \mathsf{QC}_{\{y\}}({\mathcal{O}}_{y})\cong \mathsf{QC}_{\{y\}}(\widehat{{\mathcal{O}}}_{y})$ holds (where the equivalence is given by pullback, see Thomason–Trobaugh’s theorem (Theorem 3.19)), and we assume by contradiction $N^{0}\otimes _{\mathsf{A}^{0}}\widehat{{\mathcal{O}}}_{x}=0$ . The first line of the commutative diagram contains the two face maps of the co-simplicial object $\mathsf{A}_{X,T_{\bullet }}^{\bullet }$ . Using our assumptions we therefore obtain

$$\begin{eqnarray}(N^{0}\otimes _{{\mathcal{O}}_{X}}F\otimes _{\mathsf{A}^{0}}\widehat{{\mathcal{O}}}_{y})\times _{\widehat{{\mathcal{O}}}_{y}}\mathsf{A}(x<y)\simeq N^{0}\otimes _{{\mathcal{O}}_{X}}F\otimes _{\mathsf{A}^{0}}\widehat{{\mathcal{O}}}_{x}=0.\end{eqnarray}$$

Since $N^{0}\otimes _{{\mathcal{O}}_{X}}F\in \mathsf{Perf}_{\{y\}}({\mathcal{O}}_{Y})\simeq \mathsf{Perf}_{\{y\}}(\widehat{{\mathcal{O}}}_{Y})$ (using Thomason–Trobaugh’s theorem (Theorem 3.19)), commutativity of the left-hand square implies therefore

$$\begin{eqnarray}(N^{0}\otimes _{{\mathcal{O}}_{X}}F\otimes _{\mathsf{A}^{0}}\widehat{{\mathcal{O}}}_{y})\otimes _{{\mathcal{O}}_{y}}(\widehat{{\mathcal{O}}}_{x}\otimes _{{\mathcal{O}}_{X}}{\mathcal{O}}_{y})=0.\end{eqnarray}$$

The ring homomorphism ${\mathcal{O}}_{y}\rightarrow \widehat{{\mathcal{O}}}_{x}\otimes _{{\mathcal{O}}_{X}}{\mathcal{O}}_{y}$ is the co-base change of the fully faithful ring homomorphism ${\mathcal{O}}_{x}\rightarrow \widehat{{\mathcal{O}}}_{x}$ . Therefore, we have $N^{0}\otimes _{{\mathcal{O}}_{X}}F\otimes _{\mathsf{A}^{0}}\widehat{{\mathcal{O}}}_{y}=0$ , which is a contradiction.◻

Proof. We have seen in Lemma 3.21 that there exists a closed point $x\in X$ , such that the perfect complex of $\widehat{{\mathcal{O}}_{x}}$ -modules $N^{0}\otimes _{\mathsf{A}_{X}^{0}}\widehat{{\mathcal{O}}}_{x}$ is non-zero. Using the strategy as in the proof of Lemma 3.20, we deduce the existence of a perfect complex $F\in \mathsf{Perf}_{\{x\}}(X)$ , such that $F\otimes _{{\mathcal{O}}_{X}}N^{\bullet }\neq 0$ .◻

Proof. By definition, the functor $\mathsf{A}_{X,T}(-)\simeq \mathsf{QCoh}(X)\rightarrow \mathsf{Mod}({\mathcal{O}}_{X})$ commutes with filtered colimits. Since the same is true for ${\mathcal{H}}^{i}(-)$ and $-\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{X,T}({\mathcal{O}}_{X})$ , we see that we may assume that $M$ is a perfect complex. Since the statement is local, we may assume that $M$ can be represented by a bounded complex $(V^{\bullet })$ of free sheaves of finite generation $V^{i}={\mathcal{O}}_{X}^{m_{i}}$ . In particular, ${\mathcal{H}}^{i}$ is represented as the cohomology sheaf in degree $i$ of the chain complex

$$\begin{eqnarray}\cdots \rightarrow {\mathcal{O}}_{X}^{m_{i-1}}\rightarrow {\mathcal{O}}_{X}^{m_{i}}\rightarrow {\mathcal{O}}_{X}^{m_{i+1}}\rightarrow \cdots \,.\end{eqnarray}$$

Since $\mathsf{A}_{X,T}(-)$ is exact, we obtain a chain complex

$$\begin{eqnarray}\cdots \rightarrow \mathsf{A}_{X,T}({\mathcal{O}}_{X}^{m_{i-1}})\rightarrow \mathsf{A}_{X,T}({\mathcal{O}}_{X}^{m_{i}})\rightarrow \mathsf{A}_{X,T}({\mathcal{O}}_{X}^{m_{i-1}})\rightarrow \cdots \,,\end{eqnarray}$$

and an equivalence of middle cohomology sheaves. The second chain complex can be identified with $M\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{X,T}$ . This implies the assertion.◻

Proof of Proposition 3.18.

Let $0\neq N^{\bullet }\in \mathsf{Perf}(\mathsf{A}_{X}^{\bullet })$ and $F\in \mathsf{Perf}(X)$ as in Lemma 3.22. In particular, there exists a closed point $x$ , such that $F\in \mathsf{Perf}_{\{x\}}(X)$ . By the projection formula (Lemma 3.11) we have

$$\begin{eqnarray}\int (N^{\bullet }\otimes _{{\mathcal{O}}_{X}}F)\cong F\otimes _{{\mathcal{O}}_{X}}\int N^{\bullet }.\end{eqnarray}$$

Therefore, it suffices to show that this expression is $\neq 0$ , to deduce that $\int N^{\bullet }\neq 0$ . Henceforth, we replace $N^{\bullet }$ by $N^{\bullet }\otimes F$ , that is, we have $N^{\bullet }|_{X\setminus \{x\}}=0$ . Since pullback along $\operatorname{Spec}\widehat{{\mathcal{O}}}_{x}\rightarrow X$ induces an equivalence of $\infty$ -categories (using Thomason–Trobaugh’s theorem (Theorem 3.19))

$$\begin{eqnarray}\mathsf{Perf}_{\{x\}}(X)\rightarrow \mathsf{Perf}_{\{x\}}(\widehat{{\mathcal{O}}}_{x}),\end{eqnarray}$$

one obtains $N^{0}\cong M\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{X}^{0}$ , for some $M\in \mathsf{Perf}_{\{x\}}(X)$ . Let $d_{i}:\mathsf{A}_{X}^{0}\rightarrow \mathsf{A}_{X}^{n}$ be an arbitrary boundary map, and $\unicode[STIX]{x1D719}_{i}:{\mathcal{O}}_{X}\rightarrow \mathsf{A}_{X}^{n}$ the map given by composition with the augmentation. We claim that $M\rightarrow M\otimes \mathsf{A}_{X}^{n}$ is a quasi-isomorphism of complexes of ${\mathcal{O}}_{X}$ -modules. Indeed, passing to the cohomology sheaves, we obtain using Lemma 3.23

$$\begin{eqnarray}{\mathcal{H}}^{i}(M)\rightarrow \mathsf{A}_{X}^{n}({\mathcal{H}}^{i}(M)).\end{eqnarray}$$

This is an isomorphism, since ${\mathcal{H}}^{i}(M)$ is set-theoretically supported at the closed point $x$ (Remark 1.9).

Hence, we see that $N^{\bullet }\otimes _{{\mathcal{O}}_{X}}F$ is equivalent to the constant object $M\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{X}^{\bullet }$ in $\mathsf{Perf}(\mathsf{A}_{X}^{\bullet })$ , which implies $\int N^{\bullet }\otimes _{{\mathcal{O}}_{X}}F\cong N^{0}\otimes _{{\mathcal{O}}_{X}}F\neq 0$ . ◻

Proof of Lemma 3.24.

From now on we replace $X$ by $U$ without loss of generality. Recall that the co-simplicial algebra $\mathsf{A}_{X,|Z|}^{\bullet }({\mathcal{O}}_{\unicode[STIX]{x1D702}})$ is a subobject of the co-simplicial algebra

$$\begin{eqnarray}[n]\mapsto \mathop{\prod }_{[n]\rightarrow |Z|_{{\leqslant}}}\mathsf{A}_{X,\unicode[STIX]{x1D70E}}({\mathcal{O}}_{\unicode[STIX]{x1D702}})\end{eqnarray}$$

(Lemma 1.8). For $\unicode[STIX]{x1D70E}:[n]\rightarrow |Z|_{{\leqslant}}$ we denote by $\unicode[STIX]{x1D70E}_{+}$ the map $[n+1]\rightarrow |Z|_{{\leqslant}}$ , which restricts to $\unicode[STIX]{x1D70E}$ along the map $[n]\rightarrow [n+1]$ , and sends $n+1$ to  $\unicode[STIX]{x1D702}$ . By definition, we have $\mathsf{A}_{X,\unicode[STIX]{x1D70E}}({\mathcal{O}}_{\unicode[STIX]{x1D702}})=\mathsf{A}_{X,\unicode[STIX]{x1D70E}_{+}}({\mathcal{O}}_{X})$ . Therefore, we have for $\unicode[STIX]{x1D6FC}:[n]\rightarrow [1]$ that the chain of points corresponding to $h_{n,\unicode[STIX]{x1D6FC}}(\unicode[STIX]{x1D70E})_{+}$ can be extended to the chain $\unicode[STIX]{x1D70E}_{+}$ , that is, there is a canonical inclusion

$$\begin{eqnarray}i_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FC}}:\mathsf{A}_{X,h_{n,\unicode[STIX]{x1D6FC}}(\unicode[STIX]{x1D70E})}({\mathcal{O}}_{\unicode[STIX]{x1D702}})\rightarrow \mathsf{A}_{X,\unicode[STIX]{x1D70E}}({\mathcal{O}}_{\unicode[STIX]{x1D702}}).\end{eqnarray}$$

We define the homotopy as a map

$$\begin{eqnarray}h^{n,\unicode[STIX]{x1D6FC}}:\mathop{\prod }_{\unicode[STIX]{x1D70E}:[n]\rightarrow |Z|_{{\leqslant}}}\mathsf{A}_{X,\unicode[STIX]{x1D70E}}({\mathcal{O}}_{\unicode[STIX]{x1D702}})\rightarrow \mathop{\prod }_{\unicode[STIX]{x1D70E}:[n]\rightarrow |Z|_{{\leqslant}}}\mathsf{A}_{X,\unicode[STIX]{x1D70E}}({\mathcal{O}}_{\unicode[STIX]{x1D702}}),\end{eqnarray}$$

sending $(x_{\unicode[STIX]{x1D70E}})_{\unicode[STIX]{x1D70E}:[n]\rightarrow |Z|_{{\leqslant}}}$ to $(i_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FC}}(x_{h_{n,\unicode[STIX]{x1D6FC}}(\unicode[STIX]{x1D70E})}))_{\unicode[STIX]{x1D70E}:[n]\rightarrow |Z|_{{\leqslant}}}$ . It remains to verify that the homotopy $h^{n,\unicode[STIX]{x1D6FC}}$ respects the subobject $\mathsf{A}_{X,|Z|}^{\bullet }({\mathcal{O}}_{\unicode[STIX]{x1D702}})$ .

To see this for a given $\unicode[STIX]{x1D6FC}:[n]\rightarrow [1]$ , we observe that the ring homomorphism $h^{n,\unicode[STIX]{x1D6FC}}:\mathsf{A}_{X}^{n}({\mathcal{O}}_{\unicode[STIX]{x1D702}})\rightarrow \mathsf{A}_{X}^{n}({\mathcal{O}}_{\unicode[STIX]{x1D702}})$ can be obtained using the co-simplicial structure $\mathsf{A}_{X}^{\bullet }({\mathcal{O}}_{\unicode[STIX]{x1D702}})$ . For $\unicode[STIX]{x1D6FC}:[n]\rightarrow [1]$ , we denote by $[k]=\unicode[STIX]{x1D6FC}^{-1}(0)_{+}$ the preimage of $0$ , with an additional maximal element $\unicode[STIX]{x1D702}$ adjoined (otherwise $k$ could be empty). We let $s:[k]\rightarrow [n]_{+}$ be the inclusion, which identifies $[k]\setminus \{\unicode[STIX]{x1D702}\}$ with $\unicode[STIX]{x1D6FC}^{-1}(0)$ , and preserves maxima. We define $r:[n]_{+}\rightarrow [k]$ to be the unique map in $\unicode[STIX]{x1D6E5}$ , preserving maxima, and satisfying $r\circ s=\operatorname{id}_{[k]}$ . We let $p=s\circ r:[n]_{+}\rightarrow [n]_{+}$ , and define $T=\{(x_{0}\leqslant \cdots \leqslant x_{n+1}\in |X|_{n+1}|x_{n+1}=\unicode[STIX]{x1D702})\}$ . By virtue of [Reference HuberHub91, §§ 2.2 and 2.3] we have a morphism $\mathsf{A}_{X}^{n}({\mathcal{O}}_{\unicode[STIX]{x1D702}})=\mathsf{A}_{X,T}^{n+1}({\mathcal{O}}_{X})\rightarrow \mathsf{A}_{X,T}^{n+1}({\mathcal{O}}_{X})=\mathsf{A}_{X}^{n}({\mathcal{O}}_{\unicode[STIX]{x1D702}})$ , which agrees with $h^{n,\unicode[STIX]{x1D6FC}}$ on the local factors. This shows that the homotopies $h^{n,\unicode[STIX]{x1D6FC}}$ are well-defined.

Hence, we have constructed a homotopy between $\operatorname{id}_{\mathsf{A}_{X,|Z|}^{\bullet }({\mathcal{O}}_{\unicode[STIX]{x1D702}})}$ and the composition

$$\begin{eqnarray}\mathsf{A}_{X,|Z|}^{\bullet }({\mathcal{O}}_{\unicode[STIX]{x1D702}})\rightarrow (\widehat{{\mathcal{O}}}_{\unicode[STIX]{x1D702}})^{\bullet }\rightarrow \mathsf{A}_{X,|Z|}^{\bullet }({\mathcal{O}}_{\unicode[STIX]{x1D702}}).\end{eqnarray}$$

This concludes the proof. ◻

Proof. By the projection formula (Lemma 3.11) we have $(\int M^{\bullet })_{\unicode[STIX]{x1D702}}\simeq \int (M^{\bullet }\otimes _{{\mathcal{O}}_{X}}{\mathcal{O}}_{\unicode[STIX]{x1D702}})$ . The co-simplicial object $M^{\bullet }\otimes _{{\mathcal{O}}_{X}}{\mathcal{O}}_{\unicode[STIX]{x1D702}}$ can be viewed as an $\mathsf{A}_{X,|Z|_{\bullet }}^{\bullet }({\mathcal{O}}_{\unicode[STIX]{x1D702}})$ -module by Lemma 3.14. Since the co-simplicial sheaf of algebras $\mathsf{A}_{X,|Z|_{\bullet }}^{\bullet }({\mathcal{O}}_{\unicode[STIX]{x1D702}})$ is homotopy equivalent to the constant co-simplicial sheaf of algebras $\widehat{{\mathcal{O}}}_{\unicode[STIX]{x1D702}}$ , with homotopy inverse given by the projection $\mathsf{A}_{X,|Z|_{\bullet }}^{\bullet }({\mathcal{O}}_{\unicode[STIX]{x1D702}})\rightarrow \widehat{{\mathcal{O}}}_{\unicode[STIX]{x1D702}}$ , we have an equivalence

(5) $$\begin{eqnarray}|\mathsf{Perf}(\mathsf{A}_{X,|Z|_{\bullet }}^{\bullet }({\mathcal{O}}_{\unicode[STIX]{x1D702}}))|\rightarrow \mathsf{Perf}(\widehat{{\mathcal{O}}}_{\unicode[STIX]{x1D702}}),\end{eqnarray}$$

inverse to the realisation functor given by tensoring along $\mathsf{A}_{X,|Z|_{\bullet }}^{\bullet }({\mathcal{O}}_{\unicode[STIX]{x1D702}})\rightarrow \widehat{{\mathcal{O}}}_{\unicode[STIX]{x1D702}}$ . By Corollary 3.15 we have an equivalence of functors

$$\begin{eqnarray}\int _{X,|Z|}~\circ ~(-\otimes _{\mathsf{A}_{X}^{\bullet }}\mathsf{A}_{X,|Z|}^{\bullet })\simeq \int _{X}:|\mathsf{Perf}_{|Z|}(\mathsf{A}_{X}^{\bullet })|\rightarrow \mathsf{QC}_{|Z|}(X).\end{eqnarray}$$

This implies that $(\int M^{\bullet })_{\unicode[STIX]{x1D702}}$ is a perfect $\widehat{{\mathcal{O}}}_{\unicode[STIX]{x1D702}}$ -complex. Therefore, we see that the equivalence of (5) restricted to the full subcategory $\mathsf{Perf}_{|Z|}(\mathsf{A}_{X}^{\bullet })$ lands in $\mathsf{Perf}_{\{\unicode[STIX]{x1D702}\}}(\widehat{{\mathcal{O}}}_{\unicode[STIX]{x1D702}})$ . In Theorem 3.19, we have recorded the result of Thomason–Trobaugh that this category is equivalent to $\mathsf{Perf}_{\{\unicode[STIX]{x1D702}\}}({\mathcal{O}}_{\unicode[STIX]{x1D702}})$ . This shows that $(\int M^{\bullet })_{\unicode[STIX]{x1D702}}$ is a perfect complex of ${\mathcal{O}}_{\unicode[STIX]{x1D702}}$ -modules.◻

Proof. We begin by discussing the special case $|X|=|Z|$ to which we will eventually reduce the general case.

Let $\mathsf{A}_{X,<\unicode[STIX]{x1D702}}^{0}=\prod _{x<\unicode[STIX]{x1D702}}\widehat{{\mathcal{O}}}_{x}$ , which is a factor of $\mathsf{A}_{X}^{0}=\prod _{x\in |X|}\widehat{{\mathcal{O}}}_{x}$ . By assumption, $\mathsf{A}_{X,<\unicode[STIX]{x1D702}}^{0}\otimes _{{\mathcal{O}}_{X}}{\mathcal{O}}_{\unicode[STIX]{x1D702}}$ is a factor of $\mathsf{A}_{X}^{1}$ . Via the co-simplicial structure, the second ring receives a map from $\mathsf{A}_{X,<\unicode[STIX]{x1D702}}^{0}$ and also from ${\mathcal{O}}_{\unicode[STIX]{x1D702}}$ (another factor of $\mathsf{A}_{X}^{0}$ ):

The assumption $M^{0}\otimes _{\mathsf{A}_{X}^{0}}{\mathcal{O}}_{\unicode[STIX]{x1D702}}\simeq 0$ implies therefore also $M^{1}\otimes _{\mathsf{A}_{X}^{1}}(\mathsf{A}_{X,<\unicode[STIX]{x1D702}}^{0}\otimes _{{\mathcal{O}}_{X}}{\mathcal{O}}_{\unicode[STIX]{x1D702}})\simeq 0$ .

Using that $M^{\bullet }\in |\mathsf{Perf}(\mathsf{A}_{X}^{\bullet })|$ is a cartesian co-simplicial perfect complex, we see that

$$\begin{eqnarray}0\simeq M^{0}\otimes _{\mathsf{A}_{X}^{0}}(\mathsf{A}_{X,<\unicode[STIX]{x1D702}}^{0}\otimes {\mathcal{O}}_{\unicode[STIX]{x1D702}})\simeq M^{0}\otimes _{{\mathcal{O}}_{X}}{\mathcal{O}}_{\unicode[STIX]{x1D702}},\end{eqnarray}$$

where we have used that $M^{0}\otimes _{\mathsf{A}_{X}^{0}}{\mathcal{O}}_{\unicode[STIX]{x1D702}}\simeq 0$ .

We can write ${\mathcal{O}}_{\unicode[STIX]{x1D702}}$ as a filtered colimit of rings $\mathsf{colim}_{f\in {\mathcal{O}}_{X}}{\mathcal{O}}_{X,f}$ , and hence see that

$$\begin{eqnarray}\mathsf{A}_{X,|Z|,<\unicode[STIX]{x1D702}}^{0}\otimes _{{\mathcal{O}}_{X}}{\mathcal{O}}_{\unicode[STIX]{x1D702}}\simeq \mathsf{colim}_{f\in {\mathcal{O}}_{X}}\mathsf{A}_{X,|Z|,<\unicode[STIX]{x1D702}}^{0}\otimes _{{\mathcal{O}}_{X}}{\mathcal{O}}_{X,f}\end{eqnarray}$$

is a filtered colimit of rings.

By [Reference Thomason and TrobaughTT90, Proposition 3.20], if $R=\mathsf{colim}_{i\in I}R_{i}$ is a filtered colimit of rings, and for $j\in I$ , $M_{j}\in \mathsf{Perf}(R_{j})$ , such that $M_{j}\otimes _{R_{j}}R\simeq 0$ , then there exists $i\geqslant j$ , such that $M_{j}\otimes _{R_{j}}R_{i}\simeq 0$ (in [Reference Thomason and TrobaughTT90] it is shown that the homotopy category of perfect complexes of $R$ -modules is equivalent to the filtered colimit of the homotopy category of perfect complexes of $R_{i}$ -modules).

Since our sheaves of algebras are lâche, we may apply this result, since by virtue of Corollary 2.23 the categories over perfect complexes over these sheaves of algebras are equivalent to perfect complexes over the rings of global sections. Hence, we see that there exists $f\in {\mathcal{O}}_{X}$ , such that $(M^{0}\otimes _{\mathsf{A}_{X}^{0}}\mathsf{A}_{X,<\unicode[STIX]{x1D702}}^{0})\otimes {\mathcal{O}}_{f}\simeq 0$ . Restricting both sides to the open subset $V\subset X$ defined by localisation at  $f$ , we obtain $M^{0}|_{V}\simeq 0$ , because $({\mathcal{O}}_{f})|_{V}\simeq {\mathcal{O}}$ . This concludes the proof of the case $|X|=|Z|$ .

Let $i:Z{\hookrightarrow}X$ be a closed immersion, such that its underlying closed subset is the one of interest $|Z|$ . By Lemma 1.11 we have a canonical map $\mathsf{A}_{X,|Z|}^{\bullet }\rightarrow i_{\ast }\mathsf{A}_{Z}^{\bullet }$ . Tensoring along this map produces a functor $|\mathsf{Perf}(\mathsf{A}_{X,|Z|}^{\bullet })|\rightarrow |\mathsf{Perf}(\mathsf{A}_{Z}^{\bullet })|$ .

For any $M^{\bullet }$ in $|\mathsf{Perf}(\mathsf{A}_{X,|Z|}^{\bullet })|$ , such that $M^{0}\otimes _{\mathsf{A}_{X,|Z|}^{\bullet }}\widehat{{\mathcal{O}}}_{X,\unicode[STIX]{x1D702}}\simeq 0$ , we obtain therefore by functoriality $N^{\bullet }=M^{\bullet }\otimes _{\mathsf{A}_{X,|Z|}^{\bullet }}\mathsf{A}_{X}^{\bullet }$ , which also satisfies vanishing at the generic point. Our treatment of the special case $|X|=|Z|$ can therefore be applied to  $Z$ , and we obtain an open subscheme $U\subset Z$ , such that for every $x\in U$ we have $N^{0}\otimes _{\mathsf{A}_{X}^{\bullet }}\widehat{{\mathcal{O}}}_{Z,x}\simeq 0$ .

To conclude the proof it remains to show that we also have $M^{0}\otimes _{\mathsf{A}_{X,|Z|}^{\bullet }}\widehat{{\mathcal{O}}}_{X,x}\simeq 0$ for every $x\in U$ . This is simply a consequence of Nakayama’s lemma: the map $\widehat{{\mathcal{O}}}_{X,x}\rightarrow \widehat{{\mathcal{O}}}_{Z,x}$ is a morphism of local rings inducing an isomorphism of residue field. Since a perfect complex over a local ring is acyclic if and only if it is acyclic over the residue fields, this concludes the proof.◻

Proof. Choose an isomorphism of $(\int M^{\bullet })\rightarrow \mathsf{colim}_{i\in I}N_{i}$ with a filtered colimit of perfect complexes on  $X$ . By Corollary 3.25, we know that $(\int M^{\bullet })_{\unicode[STIX]{x1D702}}$ is a perfect ${\mathcal{O}}_{\unicode[STIX]{x1D702}}$ -complex. Therefore, we see that there exists an index $j\in I$ , such that we have a factorisation

and we see that $p=s\circ r$ defines an idempotent on $(N_{j})_{\unicode[STIX]{x1D702}}$ , such that the corresponding direct summand is equivalent to $(\int M^{\bullet })_{\unicode[STIX]{x1D702}}$ . Since $N_{j}$ is perfect, there exists an affine open subset $U\subset X$ , and an idempotent $p$ on $N_{j}|_{U}$ , which extends the one over ${\mathcal{O}}_{\unicode[STIX]{x1D702}}$ . Passing to the corresponding direct summand $N_{j}^{\prime }$ of $N_{j}|_{U}$ , we see that we have constructed a direct summand $N_{j}^{\prime }$ , with a map to $(\int M^{\bullet })|_{U}$ , which induces an equivalence at $\unicode[STIX]{x1D702}$ . By adjunction, we have a morphism $N_{j}^{\prime }\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{X,|Z|}^{\bullet }\rightarrow M^{\bullet }$ , which is an equivalence after tensoring with $-\otimes _{\mathsf{A}_{X,|Z|}^{0}}{\mathcal{O}}_{\unicode[STIX]{x1D702}}$ . By Lemma 3.26 there exists an open subset $V\subset U$ , such that $N_{j}^{\prime }\otimes _{{\mathcal{O}}_{X}}\mathsf{A}_{V,|Z|\cap V}^{\bullet }\rightarrow M^{\bullet }|_{V}$ is an equivalence. This implies that $(\int M^{\bullet })|_{V}$ is equivalent to $N_{j}^{\prime }|_{V}$ , and thus is a perfect complex.◻