Lemma 30.7.2. With notation as in diagram (30.5.0.1). Assume $f : X \to S$ and $\mathcal{F}$ satisfy the hypotheses of Lemma 30.7.1. Choose a finite affine open covering $\mathcal{U} : X = \bigcup U_ i$ of $X$. There is a canonical isomorphism

\[ g^*\check{\mathcal{C}}^\bullet (\mathcal{U}, f, \mathcal{F}) \longrightarrow Rf'_*\mathcal{F}' \]

in $D^{+}(S')$. Moreover, if $S' \to S$ is affine, then in fact

\[ g^*\check{\mathcal{C}}^\bullet (\mathcal{U}, f, \mathcal{F}) = \check{\mathcal{C}}^\bullet (\mathcal{U}', f', \mathcal{F}') \]

with $\mathcal{U}' : X' = \bigcup U_ i'$ where $U_ i' = (g')^{-1}(U_ i) = U_{i, S'}$ is also affine.

**Proof.**
In fact we may define $U_ i' = (g')^{-1}(U_ i) = U_{i, S'}$ no matter whether $S'$ is affine over $S$ or not. Let $\mathcal{U}' : X' = \bigcup U_ i'$ be the induced covering of $X'$. In this case we claim that

\[ g^*\check{\mathcal{C}}^\bullet (\mathcal{U}, f, \mathcal{F}) = \check{\mathcal{C}}^\bullet (\mathcal{U}', f', \mathcal{F}') \]

with $\check{\mathcal{C}}^\bullet (\mathcal{U}', f', \mathcal{F}')$ defined in exactly the same manner as in Lemma 30.7.1. This is clear from the case of affine morphisms (Lemma 30.5.1) by working locally on $S'$. Moreover, exactly as in the proof of Lemma 30.7.1 one sees that there is an isomorphism

\[ \check{\mathcal{C}}^\bullet (\mathcal{U}', f', \mathcal{F}') \longrightarrow Rf'_*\mathcal{F}' \]

in $D^{+}(S')$ since the morphisms $U_ i' \to X'$ and $U_ i' \to S'$ are still affine (being base changes of affine morphisms). Details omitted.
$\square$

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