Lemma 75.21.1. Let $S$ be a scheme. Let

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

be a cartesian diagram of algebraic spaces over $S$. 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 (75.21.0.1) is an isomorphism,

2. for any commutative diagram

$\xymatrix{ & U \ar[d] \ar[rd]^ a \\ V' \ar[r] \ar[rd]^ c & V \ar[rd]^ b & X \ar[d]^ f \\ & Y' \ar[r]^ g & Y }$

with $a, b, c$ étale, $U, V, V'$ schemes, and with $U' = V' \times _ V U$ the equivalent conditions of Derived Categories of Schemes, Lemma 75.21.1 hold for $(U \to X)^*K$ and $(U' \to X')^*K'$, and

3. there is some diagram as in (2) with $U' \to X'$ surjective.

Proof. Observe that (1) is étale local on $X'$. Working through formal implications of what is known, we see that it suffices to prove condition (1) of this lemma is equivalent to condition (1) of Derived Categories of Schemes, Lemma 36.26.1 if $X, Y, Y', X'$ are representable by schemes $X_0, Y_0, Y'_0, X'_0$. Denote $f_0, g_0, g'_0, f'_0$ the morphisms between these schemes corresponding to $f, g, g', f'$. We may assume $K = \epsilon ^*K_0$ and $K' = \epsilon ^*K'_0$ for some objects $K_0 \in D_\mathit{QCoh}(\mathcal{O}_{X_0})$ and $K'_0 \in D_\mathit{QCoh}(\mathcal{O}_{X'_0})$, see Lemma 75.4.2. Moreover, the map $Lg^*K \to K'$ is the pullback of a map $L(g_0)^*K_0 \to K'_0$ with notation as in Remark 75.6.3. Recall that $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{X, x}$ (Properties of Spaces, Lemma 66.22.1) and that we have

$K_{\overline{x}} = K_{0, x} \otimes _{\mathcal{O}_{X, x}}^\mathbf {L} \mathcal{O}_{X, \overline{x}} \quad \text{and}\quad K'_{\overline{x}'} = K'_{0, x'} \otimes _{\mathcal{O}_{X', x'}}^\mathbf {L} \mathcal{O}_{X', \overline{x}'}$

(akin to Properties of Spaces, Lemma 66.29.4). Consider the commutative diagram

$\xymatrix{ H^ i(K_{\overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}}^\mathbf {L} \mathcal{O}_{Y', \overline{y}'}) \otimes _{(\mathcal{O}_{X, \overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{Y', \overline{y}'})} \mathcal{O}_{X', \overline{x}'} \ar[r] & H^ i(K'_{\overline{x}'}) \\ H^ i(K_{0, x} \otimes _{\mathcal{O}_{Y, y}}^\mathbf {L} \mathcal{O}_{Y', y'}) \otimes _{(\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{Y', y'})} \mathcal{O}_{X', x'} \ar[u] \ar[r] & H^ i(K'_{0, x'}) \ar[u] }$

We have to show that the lower horizontal arrow is an isomorphism if and only if the upper horizontal arrow is an isomorphism. Since $\mathcal{O}_{X', x'} \to \mathcal{O}_{X', \overline{x}'}$ is faithfully flat (More on Algebra, Lemma 15.45.1) it suffices to show that the top arrow is the base change of the bottom arrow by this map. This follows immediately from the relationships between stalks given above for the objects on the right. For the objects on the left it suffices to show that

\begin{align*} & H^ i\left( (K_{0, x} \otimes _{\mathcal{O}_{X, x}}^\mathbf {L} \mathcal{O}_{X, \overline{x}}) \otimes _{\mathcal{O}_{Y, \overline{y}}}^\mathbf {L} \mathcal{O}_{Y', \overline{y}'}\right) \\ & = H^ i(K_{0, x} \otimes _{\mathcal{O}_{Y, y}}^\mathbf {L} \mathcal{O}_{Y', y'}) \otimes _{(\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{Y', y'})} (\mathcal{O}_{X, \overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{Y', \overline{y}'}) \end{align*}

This follows from More on Algebra, Lemma 15.61.5. The flatness assumptions of this lemma hold by what was said above as well as Algebra, Lemma 10.155.12 implying that $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{Y, \overline{y}}$ and that $\mathcal{O}_{Y', \overline{y}'}$ is the strict henselization of $\mathcal{O}_{Y', y'} \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{Y, \overline{y}}$. $\square$

Comment #8718 by Keerthi Madapusi on

Looks like the reference in (2) is circular? Should probably replace 0DKG with 0DJ8?

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