This is analogous to [Theorem 2.2.3, six-I].

Lemma 21.28.6. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ be a morphism of ringed topoi. Let $\mathcal{A} \subset \textit{Mod}(\mathcal{O})$ and $\mathcal{A}' \subset \textit{Mod}(\mathcal{O}')$ be weak Serre subcategories. Assume

$f$ is flat,

$f^*$ induces an equivalence of categories $\mathcal{A}' \to \mathcal{A}$,

$\mathcal{F}' \to Rf_*f^*\mathcal{F}'$ is an isomorphism for $\mathcal{F}' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}')$,

$\mathcal{C}, \mathcal{O}, \mathcal{A}$ satisfy the assumption of Situation 21.25.1,

$\mathcal{C}', \mathcal{O}', \mathcal{A}'$ satisfy the assumption of Situation 21.25.1.

Then $f^* : D_{\mathcal{A}'}(\mathcal{O}') \to D_\mathcal {A}(\mathcal{O})$ is an equivalence of categories with quasi-inverse given by $Rf_* : D_\mathcal {A}(\mathcal{O}) \to D_{\mathcal{A}'}(\mathcal{O}')$.

**Proof.**
Since $f^*$ is exact, it is clear that $f^*$ defines a functor $f^* : D_{\mathcal{A}'}(\mathcal{O}') \to D_\mathcal {A}(\mathcal{O})$ as in the statement of the lemma and that moreover this functor commutes with the truncation functors $\tau _{\geq -n}$. We already know that $f^*$ and $Rf_*$ are quasi-inverse equivalence on the corresponding bounded below categories, see Lemma 21.28.5. By Lemma 21.25.4 with $N = 0$ we see that $Rf_*$ indeed defines a functor $Rf_* : D_\mathcal {A}(\mathcal{O}) \to D_{\mathcal{A}'}(\mathcal{O}')$ and that moreover this functor commutes with the truncation functors $\tau _{\geq -n}$. Thus for $K$ in $D_\mathcal {A}(\mathcal{O})$ the map $f^*Rf_*K \to K$ is an isomorphism as this is true on trunctions. Similarly, for $K'$ in $D_{\mathcal{A}'}(\mathcal{O}')$ the map $K' \to Rf_*f^*K'$ is an isomorphism as this is true on trunctions. This finishes the proof.
$\square$

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