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
Comments (0)