Lemma 59.88.3. 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
\[ \vcenter { \xymatrix{ X \ar[d]_ f & X' \ar[d]_{f'} \ar[l] & Y \ar[l]^ h \ar[d]^ e & Y' \ar[l]^{\pi '} \ar[d]^{e'} \\ S & S' \ar[l] & T \ar[l]_ g & T' \ar[l]_\pi } } \quad \text{and}\quad \vcenter { \xymatrix{ X' \ar[d]_{f'} & & Y' \ar[ll]^{h' = h \circ \pi '} \ar[d]^{e'} \\ S' & & T' \ar[ll]_{g' = g \circ \pi } } } \]
where all squares are cartesian, $g$ quasi-compact and quasi-separated, and $\pi $ is integral surjective. Let $\mathcal{F}$ be an abelian sheaf on $T_{\acute{e}tale}$ annihilated by $n$ and set $\mathcal{F}' = \pi ^{-1}\mathcal{F}$. 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.
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