This is a version of [Lemma 2.1.10, six-I] with slightly changed hypotheses.

Lemma 21.25.4. 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 there is an integer $N$ such that

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

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

3. $R^ pf_*\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}')$ for $p \geq 0$ and $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$,

4. $R^ pf_*\mathcal{F} = 0$ for $p > N$ and $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$,

Then for $K$ in $D_\mathcal {A}(\mathcal{O})$ we have

1. $Rf_*K$ is in $D_{\mathcal{A}'}(\mathcal{O}')$,

2. the map $H^ j(Rf_*K) \to H^ j(Rf_*(\tau _{\geq -n}K))$ is an isomorphism for $j \geq N - n$.

Proof. By Lemma 21.25.2 we have $K = R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K$. By Lemma 21.23.3 we have $Rf_*K = R\mathop{\mathrm{lim}}\nolimits Rf_*\tau _{\geq -n}K$. The complexes $Rf_*\tau _{\geq -n}K$ are bounded below. The spectral sequence

$E_2^{p, q} = R^ pf_*H^ q(\tau _{\geq -n}K)$

converging to $H^{p + q}(Rf_*\tau _{\geq -n}K)$ (Derived Categories, Lemma 13.21.3) and assumption (3) show that $Rf_*\tau _{\geq -n}K$ lies in $D^+_{\mathcal{A}'}(\mathcal{O}')$, see Homology, Lemma 12.24.11. Observe that for $m \geq n$ the map

$Rf_*(\tau _{\geq -m}K) \longrightarrow Rf_*(\tau _{\geq -n}K)$

induces an isomorphism on cohomology sheaves in degrees $j \geq -n + N$ by the spectral sequences above. Hence we may apply Lemma 21.25.3 to conclude. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).