Lemma 29.22.2. Let $f : X \to Y$ be a morphism of schemes. Assume
$f$ is quasi-compact and locally of finite presentation, and
$Y$ is quasi-compact and quasi-separated.
Then the image of every constructible subset of $X$ is constructible in $Y$.
Proof. By Properties, Lemma 28.2.5 it suffices to prove this lemma in case $Y$ is affine. In this case $X$ is quasi-compact. Hence we can write $X = U_1 \cup \ldots \cup U_ n$ with each $U_ i$ affine open in $X$. If $E \subset X$ is constructible, then each $E \cap U_ i$ is constructible too, see Topology, Lemma 5.15.4. Hence, since $f(E) = \bigcup f(E \cap U_ i)$ and since finite unions of constructible sets are constructible, this reduces us to the case where $X$ is affine. In this case the result is Algebra, Theorem 10.29.10. $\square$
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Your email address will not be published.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)