Lemma 59.86.4. Let $I$, $X_ i$, $Y_ i$, $S_ i$, $T_ i$, $f_ i$, $h_ i$, $e_ i$, $g_ i$, $X$, $Y$, $S$, $T$, $f$, $h$, $e$, $g$ be as in the statement of Lemma 59.86.3. Let $0 \in I$ and let $K_0 \in D^+(T_{0, {\acute{e}tale}})$. For $i \in I$, $i \geq 0$ denote $K_ i$ the pullback of $K_0$ to $T_ i$. Denote $K$ the pullback of $K_0$ to $T$. If $f_ i^{-1}Rg_{i, *}K_ i = Rh_{i, *}e_ i^{-1}K_ i$ for all $i \geq 0$, then $f^{-1}Rg_*K = Rh_*e^{-1}K$.
Proof. It suffices to show that the base change map $f^{-1}Rg_*K \to Rh_*e^{-1}K$ induces an isomorphism on cohomology sheaves. In other words, we have to show that $f^{-1}R^ pg_*K \to R^ ph_*e^{-1}K$ is an isomorphism for all $p \in \mathbf{Z}$ if we are given that $f_ i^{-1}R^ pg_{i, *}K_ i \to R^ ph_{i, *}e_ i^{-1}K_ i$ is an isomorphism for all $i \geq 0$ and $p \in \mathbf{Z}$. At this point we can argue exactly as in the proof of Lemma 59.86.3 replacing reference to Lemma 59.51.8 by a reference to Lemma 59.52.4. $\square$
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