Lemma 74.21.5. 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

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

$E$ 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 74.21.2 and 74.21.3 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 74.21.1. This was shown in the proof of Lemma 74.21.4.
$\square$

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