Lemma 22.35.5. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $K^\bullet$ be a complex of $\mathcal{O}$-modules. Then the functor

$- \otimes _ E^\mathbf {L} K^\bullet : D(E, \text{d}) \longrightarrow D(\mathcal{O})$

of Lemma 22.35.3 is a left adjoint of the functor

$R\mathop{\mathrm{Hom}}\nolimits (K^\bullet , -) : D(\mathcal{O}) \longrightarrow D(E, \text{d})$

of Lemma 22.32.1.

Proof. The statement means that we have

$\mathop{\mathrm{Hom}}\nolimits _{D(E, \text{d})}(M, R\mathop{\mathrm{Hom}}\nolimits (K^\bullet , L^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O})}(M \otimes ^\mathbf {L}_ E K^\bullet , L^\bullet )$

bifunctorially in $M$ and $L^\bullet$. To see this we may replace $M$ by a differential graded $E$-module $P$ with property (P). We also may replace $L^\bullet$ by a K-injective complex of $\mathcal{O}$-modules $I^\bullet$. The computation of the derived functors given in the lemmas referenced in the statement combined with Lemma 22.22.3 translates the above into

$\mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(E, \text{d})})} (P, \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(K^\bullet , I^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{O})}(P \otimes _ E K^\bullet , I^\bullet )$

where $\mathcal{B} = \text{Comp}^{dg}(\mathcal{O})$. There is an evaluation map from right to left functorial in $P$ and $I^\bullet$ (details omitted). Choose a filtration $F_\bullet$ on $P$ as in the definition of property (P). By Lemma 22.20.1 and the fact that both sides of the equation are homological functors in $P$ on $K(\text{Mod}_{(E, \text{d})})$ we reduce to the case where $P$ is replaced by the differential graded $E$-module $\bigoplus F_ iP$. Since both sides turn direct sums in the variable $P$ into direct products we reduce to the case where $P$ is one of the differential graded $E$-modules $F_ iP$. Since each $F_ iP$ has a finite filtration (given by admissible monomorphisms) whose graded pieces are graded projective $E$-modules we reduce to the case where $P$ is a graded projective $E$-module. In this case we clearly have

$\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(E, \text{d})}} (P, \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(K^\bullet , I^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{O})}(P \otimes _ E K^\bullet , I^\bullet )$

as graded $\mathbf{Z}$-modules (because this statement reduces to the case $P = E[k]$ where it is obvious). As the isomorphism is compatible with differentials we conclude. $\square$

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