The Stacks project

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. $\square$