Lemma 59.52.4. Let $I$, $g_ i : X_ i \to S_ i$, $g : X \to S$, $f_ i$, $g_ i$, $h_ i$ be as in Lemma 59.51.8. Let $0 \in I$ and $K_0 \in D^+(X_{0, {\acute{e}tale}})$. For $i \geq 0$ denote $K_ i$ the pullback of $K_0$ to $X_ i$. Denote $K$ the pullback of $K$ to $X$. Then
\[ R^ pg_*K = \mathop{\mathrm{colim}}\nolimits _{i \geq 0} h_ i^{-1}R^ pg_{i, *}K_ i \]
for all $p \in \mathbf{Z}$.
Proof.
Fix an integer $p_0 \in \mathbf{Z}$. Let $a$ be an integer such that $H^ j(K_0) = 0$ for $j < a$. We will prove the formula holds for all $p \leq p_0$ by descending induction on $a$. If $a > p_0$, then we see that the left and right hand side of the formula are zero for $p \leq p_0$ by trivial vanishing, see Derived Categories, Lemma 13.16.1. Assume $a \leq p_0$. Consider the distinguished triangle
\[ H^ a(K_0)[-a] \to K_0 \to \tau _{\geq a + 1}K_0 \]
Pulling back this distinguished triangle to $X_ i$ and $X$ gives compatible distinguished triangles for $K_ i$ and $K$. For $p \leq p_0$ we consider the commutative diagram
\[ \xymatrix{ \mathop{\mathrm{colim}}\nolimits _{i \geq 0} h_ i^{-1}R^{p - 1}g_{i, *}(\tau _{\geq a + 1}K_ i) \ar[r]_-\alpha \ar[d] & R^{p - 1}g_*(\tau _{\geq a + 1}K) \ar[d] \\ \mathop{\mathrm{colim}}\nolimits _{i \geq 0} h_ i^{-1}R^ pg_{i, *}(H^ a(K_ i)[-a]) \ar[r]_-\beta \ar[d] & R^ pg_*(H^ a(K)[-a]) \ar[d] \\ \mathop{\mathrm{colim}}\nolimits _{i \geq 0} h_ i^{-1}R^ pg_{i, *}K_ i \ar[r]_-\gamma \ar[d] & R^ pg_*K \ar[d] \\ \mathop{\mathrm{colim}}\nolimits _{i \geq 0} R^ pg_{i, *}\tau _{\geq a + 1}K_ i \ar[r]_-\delta \ar[d] & R^ pg_*\tau _{\geq a + 1}K \ar[d] \\ \mathop{\mathrm{colim}}\nolimits _{i \geq 0} R^{p + 1}g_{i, *}(H^ a(K_ i)[-a]) \ar[r]^-\epsilon & R^{p + 1}g_*(H^ a(K)[-a]) } \]
with exact columns. The arrows $\beta $ and $\epsilon $ are isomorphisms by Lemma 59.51.8. The arrows $\alpha $ and $\delta $ are isomorphisms by induction hypothesis. Hence $\gamma $ is an isomorphism as desired.
$\square$
Comments (1)
Comment #9553 by Olivier Benoist on