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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