Lemma 59.92.3. Let $f : X \to Y$ be a proper morphism of schemes. Let $g : Y' \to Y$ be a morphism of schemes. Set $X' = Y' \times _ Y X$ and denote $f' : X' \to Y'$ and $g' : X' \to X$ the projections. Let $n \geq 1$ be an integer. Let $E \in D(X_{\acute{e}tale}, \mathbf{Z}/n\mathbf{Z})$. Then the base change map (59.91.5.2) $g^{-1}Rf_*E \to Rf'_*(g')^{-1}E$ is an isomorphism.
Proof. It is enough to prove this when $Y$ and $Y'$ are quasi-compact. By Morphisms, Lemma 29.28.5 we see that the dimension of the fibres of $f : X \to Y$ and $f' : X' \to Y'$ are bounded. Thus Lemma 59.92.2 implies that
and
have finite cohomological dimension in the sense of Derived Categories, Lemma 13.32.2. Choose a K-injective complex $\mathcal{I}^\bullet $ of $\mathbf{Z}/n\mathbf{Z}$-modules each of whose terms $\mathcal{I}^ n$ is an injective sheaf of $\mathbf{Z}/n\mathbf{Z}$-modules representing $E$. See Injectives, Theorem 19.12.6. By the usual proper base change theorem we find that $R^ qf'_*(g')^{-1}\mathcal{I}^ n = 0$ for $q > 0$, see Theorem 59.91.11. Hence we conclude by Derived Categories, Lemma 13.32.2 that we may compute $Rf'_*(g')^{-1}E$ by the complex $f'_*(g')^{-1}\mathcal{I}^\bullet $. Another application of the usual proper base change theorem shows that this is equal to $g^{-1}f_*\mathcal{I}^\bullet $ as desired. $\square$
Post a comment
Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)