Lemma 46.7.2. Let $U = \mathop{\mathrm{Spec}}(A)$ be an affine scheme. The bounded below derived category $D^+(\textit{Adeq}(\mathcal{O}))$ is the localization of $K^+(\mathit{QCoh}(\mathcal{O}_ U))$ at the multiplicative subset of universal quasi-isomorphisms.

**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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)