Lemma 86.9.3. Let $X_\lambda , \lambda \in \Lambda $ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ be as in Definition 86.9.1. Then $X_\lambda \to X$ is representable and a thickening.
Proof. The statement makes sense by the discussion in Spaces, Section 64.3 and 64.5. By Lemma 86.9.2 the morphisms $X_\lambda \to X$ are representable. Given $U \to X$ where $U$ is a scheme, then the discussion following Definition 86.9.1 shows that Zariski locally on $U$ the morphism factors through some $X_\mu $ with $\lambda \leq \mu $. In this case $U \times _ X X_\lambda = U \times _{X_\mu } X_\lambda $ so that $U \times _ X X_\lambda \to U$ is a base change of the thickening $X_\lambda \to X_\mu $. $\square$
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Your email address will not be published.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)