Lemma 62.11.4. Let $f : X \to Y$ be a finite type separated morphism of quasi-compact and quasi-separated schemes. Let $\Lambda$ be a torsion ring. The functor $Rf^!$ sends $D^+(Y_{\acute{e}tale}, \Lambda )$ into $D^+(X_{\acute{e}tale}, \Lambda )$. More precisely, there exists an integer $N \geq 0$ such that if $K \in D(Y_{\acute{e}tale}, \Lambda )$ has $H^ i(K) = 0$ for $i < a$ then $H^ i(Rf^!K) = 0$ for $i < a - N$.

Proof. Let $N$ be the integer found in Lemma 62.10.2. By construction, for $K \in D(Y_{\acute{e}tale}, \Lambda )$ and $L \in \in D(X_{\acute{e}tale}, \Lambda )$ we have $\mathop{\mathrm{Hom}}\nolimits _ X(L, Rf^!K) = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_!L, K)$. Suppose $H^ i(K) = 0$ for $i < a$. Then we take $L = \tau _{\leq a - N - 1}Rf^!K$. By Lemma 62.10.2 the complex $Rf_!L$ has vanishing cohomology sheaves in degrees $\leq a - 1$. Hence $\mathop{\mathrm{Hom}}\nolimits _ Y(Rf_!L, K) = 0$ by Derived Categories, Lemma 13.27.3. Hence the canonical map $\tau _{\leq a - N - 1}Rf^!K \to Rf^!K$ is zero which implies $H^ i(Rf^!K) = 0$ for $i \leq a - N - 1$. $\square$

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).