Lemma 36.26.5. Let $f : X \to S$ be a quasi-compact and quasi-separated morphism of schemes. 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 $S$, or

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

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 $g : S' \to S$ 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

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

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 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 was shown in the proof of Lemma 36.26.4.
$\square$

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