[Expose V bis, 4.1.1, SGA4]

Theorem 20.19.2 (Proper base change). Consider a cartesian square of topological spaces

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

Assume that $f$ is proper and separated. Let $E$ be an object of $D^+(X)$. Then the base change map

$g^{-1}Rf_*E \longrightarrow Rf'_*(g')^{-1}E$

of Lemma 20.18.1 is an isomorphism in $D^+(Y')$.

Proof. Let $y' \in Y'$ be a point with image $y \in Y$. It suffices to show that the base change map induces an isomorphism on stalks at $y'$. As $f$ is proper it follows that $f'$ is proper, the fibres of $f$ and $f'$ are quasi-compact and $f$ and $f'$ are closed, see Topology, Theorem 5.17.5. Moreover $f'$ is separated by Topology, Lemma 5.4.4. Thus we can apply Lemma 20.19.1 twice to see that

$(Rf'_*(g')^{-1}E)_{y'} = R\Gamma ((f')^{-1}(y'), (g')^{-1}E|_{(f')^{-1}(y')})$

and

$(Rf_*E)_ y = R\Gamma (f^{-1}(y), E|_{f^{-1}(y)})$

The induced map of fibres $(f')^{-1}(y') \to f^{-1}(y)$ is a homeomorphism of topological spaces and the pull back of $E|_{f^{-1}(y)}$ is $(g')^{-1}E|_{(f')^{-1}(y')}$. The desired result follows. $\square$

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