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

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

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

[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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)