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

Lemma 21.25.6. Let $f : (\mathcal{C}, \mathcal{O}) \to (\mathcal{C}', \mathcal{O}')$ be a morphism of ringed sites. assume moreover there is an integer $N$ such that

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

2. $f : (\mathcal{C}, \mathcal{O}) \to (\mathcal{C}', \mathcal{O}')$ and $\mathcal{A}$ satisfy the assumption of Situation 21.25.5,

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

Then for $K$ in $D_\mathcal {A}(\mathcal{O})$ the map $H^ j(Rf_*K) \to H^ j(Rf_*(\tau _{\geq -n}K))$ is an isomorphism for $j \geq N - n$.

Proof. Let $K$ be in $D_\mathcal {A}(\mathcal{O})$. By Lemma 21.25.2 we have $K = R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K$. By Lemma 21.23.3 we have $Rf_*K = R\mathop{\mathrm{lim}}\nolimits Rf_*(\tau _{\geq -n}K)$. Let $V' \in \mathcal{B}'$ and let $\{ V'_ i \to V'\}$ be an element of $\text{Cov}_{V'}$. Then we consider

$H^ j(V'_ i, Rf_*K) = H^ j(u(V'_ i), K) \quad \text{and}\quad H^ j(V'_ i, Rf_*(\tau _{\geq -n}K)) = H^ j(u(V'_ i), \tau _{\geq -n}K)$

The assumption in Situation 21.25.5 implies that the last group is independent of $n$ for $n$ large enough depending on $j$ and $d_{V'}$. Some details omitted. We apply this for $j$ and $j - 1$ and via Lemma 21.23.2 this gives that

$H^ j(V'_ i, Rf_*K) = \mathop{\mathrm{lim}}\nolimits H^ j(V'_ i, Rf_*(\tau _{\geq -n} K))$

and the system on the right is constant for $n$ larger than a constant depending only on $d_{V'}$ and $j$. Thus Lemma 21.23.6 implies that

$H^ j(Rf_*K)(V') \longrightarrow \left(\mathop{\mathrm{lim}}\nolimits H^ j(Rf_*(\tau _{\geq -n}K))\right)(V')$

is injective. Since the elements $V' \in \mathcal{B}'$ cover every object of $\mathcal{C}'$ we conclude that the map $H^ j(Rf_*K) \to \mathop{\mathrm{lim}}\nolimits H^ j(Rf_*(\tau _{\geq -n}K))$ is injective. 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 $H^ j(Rf_*(\tau _{\geq -n}K))$ is constant for $n \geq N - j$. Hence $H^ j(Rf_*K) \to H^ j(Rf_*(\tau _{\geq -n}K))$ is injective for $j \geq N - n$.

Thus we proved the lemma with “isomorphism” in the last line of the lemma replaced by “injective”. However, now choose $j$ and $n$ with $j \geq N - n$. Then consider the distinguished triangle

$\tau _{\leq -n - 1}K \to K \to \tau _{\geq -n}K \to (\tau _{\leq -n - 1}K)$

See Derived Categories, Remark 13.12.4. Since $\tau _{\geq -n}\tau _{\leq -n -1}K = 0$, the injectivity already proven for $\tau _{-n - 1}K$ implies

$0 = H^ j(Rf_*(\tau _{\leq -n - 1}K)) = H^{j + 1}(Rf_*(\tau _{\leq -n - 1}K)) = H^{j + 2}(Rf_*(\tau _{\leq -n - 1}K)) = \ldots$

By the long exact cohomology sequence associated to the distinguished triangle

$Rf_*(\tau _{\leq -n - 1}K) \to Rf_*K \to Rf_*(\tau _{\geq -n}K) \to Rf_*(\tau _{\leq -n - 1}K)$

this implies that $H^ j(Rf_*K) \to H^ j(Rf_*(\tau _{\geq -n}K))$ is an isomorphism. $\square$

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).