The Stacks project

Proof. Follows immediately from Derived Categories, Lemma 13.17.3. $\square$

Proof. If $\varphi : \mathcal{F}^\bullet \to \mathcal{G}^\bullet $ is a morphism of complexes of quasi-coherent $\mathcal{O}_ U$-modules, then $u\varphi : u\mathcal{F}^\bullet \to u\mathcal{G}^\bullet $ is a quasi-isomorphism if and only if $\varphi $ is a universal quasi-isomorphism. Hence the collection $S$ of universal quasi-isomorphisms is a saturated multiplicative system compatible with the triangulated structure by Derived Categories, Lemma 13.5.4. Hence $S^{-1}K^+(\mathit{QCoh}(\mathcal{O}_ U))$ exists and is a triangulated category, see Derived Categories, Proposition 13.5.6. We obtain a canonical functor $can : S^{-1}K^+(\mathit{QCoh}(\mathcal{O}_ U)) \to D^{+}(\textit{Adeq}(\mathcal{O}))$ by Derived Categories, Lemma 13.5.7.

Note that, almost by definition, every adequate module on $U$ has an embedding into a quasi-coherent sheaf, see Lemma 46.5.5. Hence by Derived Categories, Lemma 13.15.5 given $\mathcal{F}^\bullet \in \mathop{\mathrm{Ob}}\nolimits (K^+(\textit{Adeq}(\mathcal{O})))$ there exists a quasi-isomorphism $\mathcal{F}^\bullet \to u\mathcal{G}^\bullet $ where $\mathcal{G}^\bullet \in \mathop{\mathrm{Ob}}\nolimits (K^+(\mathit{QCoh}(\mathcal{O}_ U)))$. This proves that $can$ is essentially surjective.

Similarly, suppose that $\mathcal{F}^\bullet $ and $\mathcal{G}^\bullet $ are bounded below complexes of quasi-coherent $\mathcal{O}_ U$-modules. A morphism in $D^+(\textit{Adeq}(\mathcal{O}))$ between these consists of a pair $f : u\mathcal{F}^\bullet \to \mathcal{H}^\bullet $ and $s : u\mathcal{G}^\bullet \to \mathcal{H}^\bullet $ where $s$ is a quasi-isomorphism. Pick a quasi-isomorphism $s' : \mathcal{H}^\bullet \to u\mathcal{E}^\bullet $. Then we see that $s' \circ f : \mathcal{F} \to \mathcal{E}^\bullet $ and the universal quasi-isomorphism $s' \circ s : \mathcal{G}^\bullet \to \mathcal{E}^\bullet $ give a morphism in $S^{-1}K^{+}(\mathit{QCoh}(\mathcal{O}_ U))$ mapping to the given morphism. This proves the "fully" part of full faithfulness. Faithfulness is proved similarly. $\square$

Proof. By Topologies, Lemma 34.7.11 (and similarly for the other topologies) we may work with $\mathcal{O}$-modules on $(\textit{Aff}/U)_\tau $. Denote $\mathcal{P}$ the category of module-valued functors on $\textit{Alg}_ A$ and $\mathcal{A}$ the category of adequate functors on $\textit{Alg}_ A$. Denote $i : \mathcal{A} \to \mathcal{P}$ the inclusion functor. Denote $Q : \mathcal{P} \to \mathcal{A}$ the construction of Lemma 46.4.1. We have the commutative diagram

The left vertical equality is Lemma 46.5.3 and the right vertical equality was explained in Section 46.3. Define $A(\mathcal{F}) = Q(j(\mathcal{F}))$. Since $j$ is fully faithful it follows immediately that $A$ is a right adjoint of the inclusion functor $k$. Also, since $k$ is fully faithful too, the final assertion follows formally. $\square$

Proof. With notation as in the proof of Lemma 46.7.3 choose an injective resolution $k(\mathcal{F}) \to \mathcal{I}^\bullet $ in the category of $\mathcal{O}$-modules on $(\textit{Aff}/U)_\tau $. By Cohomology on Sites, Lemmas 21.12.2 and Lemma 46.5.8 the complex $j(\mathcal{I}^\bullet )$ is exact. On the other hand, each $j(\mathcal{I}^ n)$ is an injective object of the category of presheaves of modules by Cohomology on Sites, Lemma 21.12.1. It follows that $R^ pA(\mathcal{F}) = R^ pQ(j(k(\mathcal{F})))$. Hence the result now follows from Lemma 46.4.10. $\square$

Proof. Let $A : \textit{Mod}(\mathcal{O}) \to \textit{Adeq}(\mathcal{O})$ be the right adjoint to the inclusion functor constructed in Lemma 46.7.3. Since $A$ is left exact and since $\textit{Mod}(\mathcal{O})$ has enough injectives, $A$ has a right derived functor $RA : D^+_{\textit{Adeq}}(\mathcal{O}) \to D^+(\textit{Adeq}(\mathcal{O}))$. We claim that $RA$ is a quasi-inverse to (46.7.5.1). To see this the key fact is that if $\mathcal{F}$ is an adequate module, then the adjunction map $\mathcal{F} \to RA(\mathcal{F})$ is a quasi-isomorphism by Lemma 46.7.4.

Namely, to prove the lemma in full it suffices to show:

1. Given $\mathcal{F}^\bullet \in K^+(\textit{Adeq}(\mathcal{O}))$ the canonical map $\mathcal{F}^\bullet \to RA(\mathcal{F}^\bullet )$ is a quasi-isomorphism, and

2. given $\mathcal{G}^\bullet \in K^+(\textit{Mod}(\mathcal{O}))$ the canonical map $RA(\mathcal{G}^\bullet ) \to \mathcal{G}^\bullet $ is a quasi-isomorphism.

Both (1) and (2) follow from the key fact via a spectral sequence argument using one of the spectral sequences of Derived Categories, Lemma 13.21.3. Some details omitted. $\square$

Proof. By definition these ext groups are computed as hom sets in the derived category. Hence this follows immediately from Lemma 46.7.5. $\square$