Lemma 75.21.5. Let $S$ be a scheme. Let $f : X \to Y$ be a quasi-compact and quasi-separated morphism of algebraic spaces over $S$. Let $E$ be an object of $D(\mathcal{O}_ X)$. Let $\mathcal{G}^\bullet$ be a complex of quasi-coherent $\mathcal{O}_ X$-modules. If

1. $E$ is perfect, $\mathcal{G}^\bullet$ is a bounded above, and $\mathcal{G}^ n$ is flat over $Y$, or

2. $E$ is pseudo-coherent, $\mathcal{G}^\bullet$ is bounded, and $\mathcal{G}^ n$ is flat over $Y$,

then formation of

$Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{G}^\bullet )$

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

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

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

$Lg^*Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{G}^\bullet ) \longrightarrow R(f')_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, (g')^*\mathcal{G}^\bullet )$

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 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 was shown in the proof of Lemma 75.21.4. $\square$

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