Lemma 76.49.6. Consider a commutative diagram
of algebraic spaces. Assume that
$p$ is locally of finite presentation,
$p$ is flat,
$p$ is universally closed,
$q$ is locally of finite type,
$q$ is closed, and
$q$ is separated.
Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is an isomorphism.
Proof. By Lemma 76.49.5 there exists an open subspace $W_1 \subset Z$ such that $f_{Z'}$ is surjective and flat if and only if $Z' \to Z$ factors through $W_1$. By Lemma 76.49.3 there exists an open subspace $W_2 \subset Z$ such that $f_{Z'}$ is a closed immersion if and only if $Z' \to Z$ factors through $W_2$. We claim that $W = W_1 \cap W_2$ works. Certainly, if $f_{Z'}$ is an isomorphism, then $Z' \to Z$ factors through $W$. Hence it suffices to show that $f_ W$ is an isomorphism. By construction $f_ W$ is a surjective flat closed immersion. In particular $f_ W$ is representable. Since a surjective flat closed immersion of schemes is an isomorphism (see Morphisms, Lemma 29.26.1) we win. (Note that actually $f_ W$ is locally of finite presentation, whence open, so you can avoid the use of this lemma if you like.) $\square$
Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)