Lemma 69.4.1. Let $S$ be a scheme. Let $f : X \to Y$ be an integral (for example finite) morphism of algebraic spaces. Then $f_* : \textit{Ab}(X_{\acute{e}tale}) \to \textit{Ab}(Y_{\acute{e}tale})$ is an exact functor and $R^ pf_* = 0$ for $p > 0$.
Proof. By Properties of Spaces, Lemma 66.18.12 we may compute the higher direct images on an étale cover of $Y$. Hence we may assume $Y$ is a scheme. This implies that $X$ is a scheme (Morphisms of Spaces, Lemma 67.45.3). In this case we may apply Étale Cohomology, Lemma 59.43.5. For the finite case the reader may wish to consult the less technical Étale Cohomology, Proposition 59.55.2. $\square$
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Your email address will not be published.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)