Lemma 52.6.8. Let $(\mathcal{C}, \mathcal{O})$ be a ringed on a site. Let $f_1, \ldots , f_ r$ be global sections of $\mathcal{O}$. Let $\mathcal{I} \subset \mathcal{O}$ be the ideal sheaf generated by $f_1, \ldots , f_ r$. Then the inclusion functor $D_{comp}(\mathcal{O}) \to D(\mathcal{O})$ has a left adjoint, i.e., given any object $K$ of $D(\mathcal{O})$ there exists a map $K \to K^\wedge $ with $K^\wedge $ in $D_{comp}(\mathcal{O})$ such that the map

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O})}(K^\wedge , E) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O})}(K, E) \]

is bijective whenever $E$ is in $D_{comp}(\mathcal{O})$. In fact we have

\[ K^\wedge = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O} (\mathcal{O} \to \prod \nolimits _{i_0} \mathcal{O}_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} \mathcal{O}_{f_{i_0}f_{i_1}} \to \ldots \to \mathcal{O}_{f_1\ldots f_ r}, K) \]

functorially in $K$.

**Proof.**
Define $K^\wedge $ by the last displayed formula of the lemma. There is a map of complexes

\[ (\mathcal{O} \to \prod \nolimits _{i_0} \mathcal{O}_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} \mathcal{O}_{f_{i_0}f_{i_1}} \to \ldots \to \mathcal{O}_{f_1\ldots f_ r}) \longrightarrow \mathcal{O} \]

which induces a map $K \to K^\wedge $. It suffices to prove that $K^\wedge $ is derived complete and that $K \to K^\wedge $ is an isomorphism if $K$ is derived complete.

Let $f$ be a global section of $\mathcal{O}$. By Lemma 52.6.1 the object $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ f, K^\wedge )$ is equal to

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}( (\mathcal{O}_ f \to \prod \nolimits _{i_0} \mathcal{O}_{ff_{i_0}} \to \prod \nolimits _{i_0 < i_1} \mathcal{O}_{ff_{i_0}f_{i_1}} \to \ldots \to \mathcal{O}_{ff_1\ldots f_ r}), K) \]

If $f = f_ i$ for some $i$, then $f_1, \ldots , f_ r$ generate the unit ideal in $\mathcal{O}_ f$, hence the extended alternating Čech complex

\[ \mathcal{O}_ f \to \prod \nolimits _{i_0} \mathcal{O}_{ff_{i_0}} \to \prod \nolimits _{i_0 < i_1} \mathcal{O}_{ff_{i_0}f_{i_1}} \to \ldots \to \mathcal{O}_{ff_1\ldots f_ r} \]

is zero (even homotopic to zero). In this way we see that $K^\wedge $ is derived complete.

If $K$ is derived complete, then $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ f, K)$ is zero for all $f = f_{i_0} \ldots f_{i_ p}$, $p \geq 0$. Thus $K \to K^\wedge $ is an isomorphism in $D(\mathcal{O})$.
$\square$

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