Lemma 59.88.4. With $f : X \to S$ and $n$ as in Remark 59.88.1 assume for some $q \geq 1$ we have $BC(f, n, q - 1)$. Consider commutative diagrams

\[ \vcenter { \xymatrix{ X \ar[d]_ f & X' \ar[d]_{f'} \ar[l] & X'' \ar[l]^{\pi '} \ar[d]_{f''} & Y \ar[l]^{h'} \ar[d]^ e \\ S & S' \ar[l] & S'' \ar[l]_\pi & T \ar[l]_{g'} } } \quad \text{and}\quad \vcenter { \xymatrix{ X' \ar[d]_{f'} & & Y \ar[ll]^{h = h' \circ \pi '} \ar[d]^ e \\ S' & & T \ar[ll]_{g = g' \circ \pi } } } \]

where all squares are cartesian, $g'$ quasi-compact and quasi-separated, and $\pi $ is integral. Let $\mathcal{F}$ be an abelian sheaf on $T_{\acute{e}tale}$ annihilated by $n$. If the base change map

\[ (f')^{-1}R^ qg_*\mathcal{F} \longrightarrow R^ qh_*e^{-1}\mathcal{F} \]

is an isomorphism, then the base change map $(f'')^{-1}R^ qg'_*\mathcal{F} \to R^ qh'_*e^{-1}\mathcal{F}$ is an isomorphism.

**Proof.**
Since $\pi $ and $\pi '$ are integral we have $R\pi _* = \pi _*$ and $R\pi '_* = \pi '_*$, see Lemma 59.43.5. We also have $(f')^{-1}\pi _* = \pi '_*(f'')^{-1}$. Thus we see that $\pi '_*(f'')^{-1}R^ qg'_*\mathcal{F} = (f')^{-1}R^ qg_*\mathcal{F}$ and $\pi '_*R^ qh'_*e^{-1}\mathcal{F} = R^ qh_*e^{-1}\mathcal{F}$. Thus the assumption means that our map becomes an isomorphism after applying the functor $\pi '_*$. Hence we see that it is an isomorphism by Lemma 59.43.5.
$\square$

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