Lemma 75.21.4. Let S be a scheme. Let f : X \to Y be a quasi-compact and quasi-separated morphism of algebraic spaces over S. Let E \in D_\mathit{QCoh}(\mathcal{O}_ X). Let \mathcal{G}^\bullet be a bounded above complex of quasi-coherent \mathcal{O}_ X-modules flat over Y. Then formation of
Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{G}^\bullet )
commutes with arbitrary base change (see proof for precise statement).
Proof.
The statement means the following. Let g : Y' \to Y be a morphism of algebraic spaces and consider the base change diagram
\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y }
in other words X' = Y' \times _ Y X. The lemma asserts that
Lg^*Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{G}^\bullet ) \longrightarrow Rf'_*(L(g')^*E \otimes ^\mathbf {L}_{\mathcal{O}_{X'}} (g')^*\mathcal{G}^\bullet )
is an isomorphism. Observe that on the right hand side we do not use derived pullback on \mathcal{G}^\bullet . To prove this, we apply Lemmas 75.21.2 and 75.21.3 to see that it suffices to prove the canonical map
L(g')^*\mathcal{G}^\bullet \to (g')^*\mathcal{G}^\bullet
satisfies the equivalent conditions of Lemma 75.21.1. This follows by checking the condition on stalks, where it immediately follows from the fact that \mathcal{G}^\bullet _{\overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{Y', \overline{y}'} computes the derived tensor product by our assumptions on the complex \mathcal{G}^\bullet .
\square
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