Lemma 59.86.1. Consider a cartesian diagram of schemes
\xymatrix{ X \ar[d]_ f & Y \ar[l]^ h \ar[d]^ e \\ S & T \ar[l]_ g }
Let \{ U_ i \to X\} be an étale covering such that U_ i \to S factors as U_ i \to V_ i \to S with V_ i \to S étale and consider the cartesian diagrams
\xymatrix{ U_ i \ar[d]_{f_ i} & U_ i \times _ X Y \ar[l]^{h_ i} \ar[d]^{e_ i} \\ V_ i & V_ i \times _ S T \ar[l]_{g_ i} }
Let \mathcal{F} be a sheaf on T_{\acute{e}tale}. Let K in D(T_{\acute{e}tale}). Set K_ i = K|_{V_ i \times _ S T} and \mathcal{F}_ i = \mathcal{F}|_{V_ i \times _ S T}.
If f_ i^{-1}g_{i, *}\mathcal{F}_ i = h_{i, *}e_ i^{-1}\mathcal{F}_ i for all i, then f^{-1}g_*\mathcal{F} = h_*e^{-1}\mathcal{F}.
If f_ i^{-1}Rg_{i, *}K_ i = Rh_{i, *}e_ i^{-1}K_ i for all i, then f^{-1}Rg_*K = Rh_*e^{-1}K.
If \mathcal{F} is an abelian sheaf and f_ i^{-1}R^ qg_{i, *}\mathcal{F}_ i = R^ qh_{i, *}e_ i^{-1}\mathcal{F}_ i for all i, then f^{-1}R^ qg_*\mathcal{F} = R^ qh_*e^{-1}\mathcal{F}.
Proof.
Proof of (1). First we observe that
(f^{-1}g_*\mathcal{F})|_{U_ i} = f_ i^{-1}(g_*\mathcal{F}|_{V_ i}) = f_ i^{-1}g_{i, *}\mathcal{F}_ i
The first equality because U_ i \to X \to S is equal to U_ i \to V_ i \to S and the second equality because g_*\mathcal{F}|_{V_ i} = g_{i, *}\mathcal{F}_ i by Sites, Lemma 7.28.2. Similarly we have
(h_*e^{-1}\mathcal{F})|_{U_ i} = h_{i, *}(e^{-1}\mathcal{F}|_{U_ i \times _ X Y}) = h_{i, *}e_ i^{-1}\mathcal{F}_ i
Thus if the base change maps f_ i^{-1}g_{i, *}\mathcal{F}_ i \to h_{i, *}e_ i^{-1}\mathcal{F}_ i are isomorphisms for all i, then the base change map f^{-1}g_*\mathcal{F} \to h_*e^{-1}\mathcal{F} restricts to an isomorphism over U_ i for all i and we conclude it is an isomorphism as \{ U_ i \to X\} is an étale covering.
For the other two statements we replace the appeal to Sites, Lemma 7.28.2 by an appeal to Cohomology on Sites, Lemma 21.20.4.
\square
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