Lemma 59.52.5. Let $I$, $g_ i : X_ i \to S_ i$, $g : X \to S$, $f_{ii'}$, $f_ i$, $g_ i$, $h_ i$ be as in Lemma 59.51.8. Let $\mathcal{F}_ i^\bullet $ be a complex of abelian sheaves on $X_{i, {\acute{e}tale}}$. Let $\varphi _{i'i} : f_{i'i}^{-1}\mathcal{F}_ i^\bullet \to \mathcal{F}_{i'}^\bullet $ be a map of complexes on $X_{i, {\acute{e}tale}}$ such that $\varphi _{i''i} = \varphi _{i''i'} \circ f_{i'' i'}^{-1}\varphi _{i'i}$ whenever $i'' \geq i' \geq i$. Assume there is an integer $a$ such that $\mathcal{F}_ i^ n = 0$ for $n < a$ and all $i \in I$. Then
\[ R^ pg_*(\mathop{\mathrm{colim}}\nolimits f_ i^{-1}\mathcal{F}_ i^\bullet ) = \mathop{\mathrm{colim}}\nolimits _{i \geq 0} h_ i^{-1}R^ pg_{i, *}\mathcal{F}_ i^\bullet \]
for all $p \in \mathbf{Z}$.
Proof.
This is a consequence of Lemma 59.51.8. Set $\mathcal{F}^\bullet = \mathop{\mathrm{colim}}\nolimits f_ i^{-1}\mathcal{F}_ i^\bullet $. The lemma tells us that
\[ \mathop{\mathrm{colim}}\nolimits _{i \in I} h_ i^{-1}R^ pg_{i, *}\mathcal{F}_ i^ n = R^ pg_*\mathcal{F}^ n \]
for all $n, p \in \mathbf{Z}$. Let us use the spectral sequences
\[ E_{1, i}^{s, t} = R^ tg_{i, *}\mathcal{F}_ i^ s \Rightarrow R^{s + t}g_{i, *}\mathcal{F}_ i^\bullet \]
and
\[ E_1^{s, t} = R^ tg_*\mathcal{F}^ s \Rightarrow R^{s + t}g_*\mathcal{F}^\bullet \]
of Derived Categories, Lemma 13.21.3. Since $\mathcal{F}_ i^ n = 0$ for $n < a$ (with $a$ independent of $i$) we see that only a fixed finite number of terms $E_{1, i}^{s, t}$ (independent of $i$) and $E_1^{s, t}$ contribute and $E_1^{s, t} = \mathop{\mathrm{colim}}\nolimits E_{i, i}^{s, t}$. This implies what we want. Some details omitted. (There is an alternative argument using “stupid” truncations of complexes which avoids using spectral sequences.)
$\square$
Comments (0)