Theorem 20.18.2 (Proper base change). Consider a cartesian square of topological spaces
Assume that f is proper. Let E be an object of D^+(X). Then the base change map
of Lemma 20.17.1 is an isomorphism in D^+(Y').
[Expose V bis, 4.1.1, SGA4]
Theorem 20.18.2 (Proper base change). Consider a cartesian square of topological spaces
Assume that f is proper. Let E be an object of D^+(X). Then the base change map
of Lemma 20.17.1 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 and Lemma 5.4.4. Thus we can apply Lemma 20.18.1 twice to see that
and
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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