Lemma 76.49.3. Consider a commutative diagram
of algebraic spaces. Assume that
$p$ is locally of finite type,
$p$ is universally closed, and
$q : Y \to Z$ 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 a closed immersion.
Proof. We will use the characterization of closed immersions as universally closed, unramified, and universally injective morphisms, see Lemma 76.14.9. First, note that since $p$ is universally closed and $q$ is separated, we see that $f$ is universally closed, see Morphisms of Spaces, Lemma 67.40.6. It follows that any base change of $f$ is universally closed, see Morphisms of Spaces, Lemma 67.9.3. Thus to finish the proof of the lemma it suffices to prove that the assumptions of Lemma 76.49.2 are satisfied. The projection $\text{pr}_0 : X \times _ Y X \to X$ is universally closed as a base change of $f$, see Morphisms of Spaces, Lemma 67.9.3. Hence $X \times _ Y X \to Z$ is universally closed as a composition of universally closed morphisms (see Morphisms of Spaces, Lemma 67.9.4). This finishes the proof of the lemma. $\square$
Your email address will not be published.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)