Lemma 15.91.18. In Situation 15.91.15. The functor which sends $K \in D(A)$ to the derived limit $K' = R\mathop{\mathrm{lim}}\nolimits ( K \otimes _ A^\mathbf {L} K_ n^\bullet )$ is the left adjoint to the inclusion functor $D_{comp}(A) \to D(A)$ constructed in Lemma 15.91.10.

First proof. The assignment $K \leadsto K'$ is a functor and $K'$ is derived complete with respect to $I$ by Lemma 15.91.16. By a formal argument (omitted) we see that it suffices to show $K \to K'$ is an isomorphism if $K$ is derived complete with respect to $I$. This is Lemma 15.91.17. $\square$

Second proof. Denote $K \mapsto K^\wedge$ the adjoint constructed in Lemma 15.91.10. By that lemma we have

$K^\wedge = R\mathop{\mathrm{Hom}}\nolimits \left((A \to \prod \nolimits _{i_0} A_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_ r}), K\right)$

In Lemma 15.29.6 we have seen that the extended alternating Čech complex

$A \to \prod \nolimits _{i_0} A_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_ r}$

is a colimit of the Koszul complexes $K^ n = K(A, f_1^ n, \ldots , f_ r^ n)$ sitting in degrees $0, \ldots , r$. Note that $K^ n$ is a finite chain complex of finite free $A$-modules with dual (as in Lemma 15.74.15) $R\mathop{\mathrm{Hom}}\nolimits _ A(K^ n, A) = K_ n$ where $K_ n$ is the Koszul cochain complex sitting in degrees $-r, \ldots , 0$ (as usual). Thus it suffices to show that

$R\mathop{\mathrm{Hom}}\nolimits _ A(\text{hocolim} K^ n, K) = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} K_ n)$

This follows from Lemma 15.74.16. $\square$

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