Lemma 36.26.4. Let $f : X \to S$ be a quasi-compact and quasi-separated morphism of schemes. 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 $S$. 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 $g : S' \to S$ be a morphism of schemes and consider the base change diagram

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

in other words $X' = S' \times _ S X$. The lemma asserts that

\[ Lg^*Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{G}^\bullet ) \longrightarrow Rf'_*\left( L(g')^*E \otimes ^\mathbf {L}_{\mathcal{O}_{X'}} (g')^*\mathcal{G}^\bullet \right) \]

is an isomorphism. Observe that on the right hand side we do **not** use the derived pullback on $\mathcal{G}^\bullet $. To prove this, we apply Lemmas 36.26.2 and 36.26.3 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 36.26.1. This follows by checking the condition on stalks, where it immediately follows from the fact that $\mathcal{G}^\bullet _ x \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{S', s'}$ computes the derived tensor product by our assumptions on the complex $\mathcal{G}^\bullet $.
$\square$

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