Lemma 61.20.2. Let $\Lambda$ be a Noetherian ring. Let $I \subset \Lambda$ be an ideal. Let $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ be a morphism of topoi. Then

1. $Rf_*$ sends $D_{comp}(\mathcal{D}, \Lambda )$ into $D_{comp}(\mathcal{C}, \Lambda )$,

2. the map $Rf_* : D_{comp}(\mathcal{D}, \Lambda ) \to D_{comp}(\mathcal{C}, \Lambda )$ has a left adjoint $Lf_{comp}^* : D_{comp}(\mathcal{C}, \Lambda ) \to D_{comp}(\mathcal{D}, \Lambda )$ which is $Lf^*$ followed by derived completion,

3. $Rf_*$ commutes with derived completion,

4. for $K$ in $D_{comp}(\mathcal{D}, \Lambda )$ we have $Rf_*K = R\mathop{\mathrm{lim}}\nolimits Rf_*(K \otimes ^\mathbf {L}_\Lambda \underline{\Lambda /I^ n})$.

5. for $M$ in $D_{comp}(\mathcal{C}, \Lambda )$ we have $Lf^*_{comp}M = R\mathop{\mathrm{lim}}\nolimits Lf^*(M \otimes ^\mathbf {L}_\Lambda \underline{\Lambda /I^ n})$.

Proof. We have seen (1) and (2) in Algebraic and Formal Geometry, Lemma 52.6.18. Part (3) follows from Algebraic and Formal Geometry, Lemma 52.6.19. For (4) let $K$ be derived complete. Then

$Rf_*K = Rf_*( R\mathop{\mathrm{lim}}\nolimits K \otimes ^\mathbf {L}_\Lambda \underline{\Lambda /I^ n}) = R\mathop{\mathrm{lim}}\nolimits Rf_*(K \otimes ^\mathbf {L}_\Lambda \underline{\Lambda /I^ n})$

the first equality by Lemma 61.20.1 and the second because $Rf_*$ commutes with $R\mathop{\mathrm{lim}}\nolimits$ (Cohomology on Sites, Lemma 21.23.3). This proves (4). To prove (5), by Lemma 61.20.1 we have

$Lf_{comp}^*M = R\mathop{\mathrm{lim}}\nolimits ( Lf^*M \otimes _\Lambda ^\mathbf {L} \underline{\Lambda /I^ n})$

Since $Lf^*$ commutes with derived tensor product by Cohomology on Sites, Lemma 21.18.4 and since $Lf^*\underline{\Lambda /I^ n} = \underline{\Lambda /I^ n}$ we get (5). $\square$

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