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.

Proof. Observe that $\mathcal{F} \to \pi _*\pi ^{-1}\mathcal{F}'$ is injective as $\pi$ is surjective (check on stalks). Thus by Lemma 59.88.2 we see that it suffices to show that the base change map

$(f')^{-1}R^ qg_*\pi _*\mathcal{F}' \longrightarrow R^ qh_*e^{-1}\pi _*\mathcal{F}'$

is an isomorphism. This follows from the assumption because we have $R^ qg_*\pi _*\mathcal{F}' = R^ qg'_*\mathcal{F}'$, we have $e^{-1}\pi _*\mathcal{F}' =\pi '_*(e')^{-1}\mathcal{F}'$, and we have $R^ qh_*\pi '_*(e')^{-1}\mathcal{F}' = R^ qh'_*(e')^{-1}\mathcal{F}'$. This follows from Lemmas 59.55.4 and 59.43.5 and the relative leray spectral sequence (Cohomology on Sites, Lemma 21.14.7). $\square$

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