The Stacks project

Lemma 59.88.2. With $f : X \to S$ and $n$ as in Remark 59.88.1 assume for some $q \geq 1$ we have $BC(f, n, q - 1)$. Then for every commutative diagram

\[ \xymatrix{ X \ar[d]_ f & X' \ar[l] \ar[d]_{f'} & Y \ar[l]^ h \ar[d]^ e \\ S & S' \ar[l] & T \ar[l]_ g } \]

with $X' = X \times _ S S'$ and $Y = X' \times _{S'} T$ and $g$ quasi-compact and quasi-separated, and every abelian sheaf $\mathcal{F}$ on $T_{\acute{e}tale}$ annihilated by $n$

  1. the base change map $(f')^{-1}R^ qg_*\mathcal{F}\to R^ qh_*e^{-1}\mathcal{F}$ is injective,

  2. if $\mathcal{F} \subset \mathcal{G}$ where $\mathcal{G}$ on $T_{\acute{e}tale}$ is annihilated by $n$, then

    \[ \mathop{\mathrm{Coker}}\left( (f')^{-1}R^ qg_*\mathcal{F}\to R^ qh_*e^{-1}\mathcal{F} \right) \subset \mathop{\mathrm{Coker}}\left( (f')^{-1}R^ qg_*\mathcal{G}\to R^ qh_*e^{-1}\mathcal{G} \right) \]

  3. if in (2) the sheaf $\mathcal{G}$ is an injective sheaf of $\mathbf{Z}/n\mathbf{Z}$-modules, then

    \[ \mathop{\mathrm{Coker}}\left((f')^{-1}R^ qg_*\mathcal{F}\to R^ qh_*e^{-1}\mathcal{F} \right) \subset R^ qh_*e^{-1}\mathcal{G} \]

Proof. Choose a short exact sequence $0 \to \mathcal{F} \to \mathcal{I} \to \mathcal{Q} \to 0$ where $\mathcal{I}$ is an injective sheaf of $\mathbf{Z}/n\mathbf{Z}$-modules. Consider the induced diagram

\[ \xymatrix{ (f')^{-1}R^{q - 1}g_*\mathcal{I} \ar[d]_{\cong } \ar[r] & (f')^{-1}R^{q - 1}g_*\mathcal{Q} \ar[d]_{\cong } \ar[r] & (f')^{-1}R^ qg_*\mathcal{F} \ar[d] \ar[r] & 0 \ar[d] \\ R^{q - 1}h_*e^{-1}\mathcal{I} \ar[r] & R^{q - 1}h_*e^{-1}\mathcal{Q} \ar[r] & R^ qh_*e^{-1}\mathcal{F} \ar[r] & R^ qh_*e^{-1}\mathcal{I} } \]

with exact rows. We have the zero in the right upper corner as $\mathcal{I}$ is injective. The left two vertical arrows are isomorphisms by $BC(f, n, q - 1)$. We conclude that part (1) holds. The above also shows that

\[ \mathop{\mathrm{Coker}}\left( (f')^{-1}R^ qg_*\mathcal{F}\to R^ qh_*e^{-1}\mathcal{F} \right) \subset R^ qh_*e^{-1}\mathcal{I} \]

hence part (3) holds. To prove (2) choose $\mathcal{F} \subset \mathcal{G} \subset \mathcal{I}$. $\square$

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