$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S }$

be a cartesian diagram of schemes. Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $L(g')^*K \to K'$ be a map in $D_\mathit{QCoh}(\mathcal{O}_{X'})$. The following are equivalent

1. for any $x' \in X'$ and $i \in \mathbf{Z}$ the map (36.26.0.1) is an isomorphism,

2. for $U \subset X$, $V' \subset S'$ affine open both mapping into the affine open $V \subset S$ with $U' = V' \times _ V U$ the composition

$R\Gamma (U, K) \otimes _{\mathcal{O}_ S(U)}^\mathbf {L} \mathcal{O}_{S'}(V') \to R\Gamma (U, K) \otimes _{\mathcal{O}_ X(U)}^\mathbf {L} \mathcal{O}_{X'}(U') \to R\Gamma (U', K')$

is an isomorphism in $D(\mathcal{O}_{S'}(V'))$, and

3. there is a set $I$ of quadruples $U_ i, V_ i', V_ i, U_ i'$, $i \in I$ as in (2) with $X' = \bigcup U'_ i$.

Proof. The second arrow in (2) comes from the equality

$R\Gamma (U, K) \otimes _{\mathcal{O}_ X(U)}^\mathbf {L} \mathcal{O}_{X'}(U') = R\Gamma (U', L(g')^*K)$

of Lemma 36.3.8 and the given arrow $L(g')^*K \to K'$. The first arrow of (2) is More on Algebra, Equation (15.61.0.1). It is clear that (2) implies (3). Observe that (1) is local on $X'$. Therefore it suffices to show that if $X$, $S$, $S'$, $X'$ are affine, then (1) is equivalent to the condition that

$R\Gamma (X, K) \otimes _{\mathcal{O}_ S(S)}^\mathbf {L} \mathcal{O}_{S'}(S') \to R\Gamma (X, K) \otimes _{\mathcal{O}_ X(X)}^\mathbf {L} \mathcal{O}_{X'}(X') \to R\Gamma (X', K')$

is an isomorphism in $D(\mathcal{O}_{S'}(S'))$. Say $S = \mathop{\mathrm{Spec}}(R)$, $X = \mathop{\mathrm{Spec}}(A)$, $S' = \mathop{\mathrm{Spec}}(R')$, $X' = \mathop{\mathrm{Spec}}(A')$, $K$ corresponds to the complex $M^\bullet$ of $A$-modules, and $K'$ corresponds to the complex $N^\bullet$ of $A'$-modules. Note that $A' = A \otimes _ R R'$. The condition above is that the composition

$M^\bullet \otimes _ R^\mathbf {L} R' \to M^\bullet \otimes _ A^\mathbf {L} A' \to N^\bullet$

is an isomorphism in $D(R')$. Equivalently, it is that for all $i \in \mathbf{Z}$ the map

$H^ i(M^\bullet \otimes _ R^\mathbf {L} R') \to H^ i(M^\bullet \otimes _ A^\mathbf {L} A') \to H^ i(N^\bullet )$

is an isomorphism. Observe that this is a map of $A \otimes _ R R'$-modules, i.e., of $A'$-modules. On the other hand, (1) is the requirement that for compatible primes $\mathfrak q' \subset A'$, $\mathfrak q \subset A$, $\mathfrak p' \subset R'$, $\mathfrak p \subset R$ the composition

$H^ i(M^\bullet _\mathfrak q \otimes _{R_\mathfrak p}^\mathbf {L} R'_{\mathfrak p'}) \otimes _{(A_\mathfrak q \otimes _{R_\mathfrak p} R'_{\mathfrak p'})} A'_{\mathfrak q'} \to H^ i(M^\bullet _{\mathfrak q} \otimes _{A_\mathfrak q}^\mathbf {L} A'_{\mathfrak q'}) \to H^ i(N^\bullet _{\mathfrak q'})$

is an isomorphism. Since

$H^ i(M^\bullet _\mathfrak q \otimes _{R_\mathfrak p}^\mathbf {L} R'_{\mathfrak p'}) \otimes _{(A_\mathfrak q \otimes _{R_\mathfrak p} R'_{\mathfrak p'})} A'_{\mathfrak q'} = H^ i(M^\bullet \otimes _ R^\mathbf {L} R') \otimes _{A'} A'_{\mathfrak q'}$

is the localization at $\mathfrak q'$, we see that these two conditions are equivalent by Algebra, Lemma 10.23.1. $\square$

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