Lemma 68.13.9. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$ be a point. Assume
$|f|(|Y|)$ is contained in $\{ x\} \subset |X|$,
$Y$ is reduced, and
$X$ is locally Noetherian.
Then $f$ factors through the residual space $Z_ x$ of $X$ at $x$.
Proof. Preliminary remark: since $Z_ x \to X$ is a monomorphism, it suffices to find a surjective étale morphism $Y' \to Y$ such that $Y' \to X$ factors through $Z_ x$. A remark here is that $Y'$ is reduced as well.
Let $U$ be an affine scheme and let $U \to X$ be an étale morphism such that $x$ is in the image of $|U| \to |X|$. Since $X$ is locally Noetherian, $U$ is a Noetherian affine scheme. By assumption (1) we see that $Y' = U \times _ X Y \to Y$ is surjective as well as étale. Denote $E \subset |U|$ the set of points mapping to $x$. There are no nontrivial specializations between the elements of $E$, see Lemma 68.7.2. The morphism $Y' \to U$ maps $|Y'|$ into $E$. By our construction of $Z_ x$ in the proof of Lemma 68.13.5 we know that $\coprod _{u \in E} u \to X$ factors through $Z_ x$. Hence it suffices to prove that $Y' \to U$ factors through $\coprod _{u \in E} u \to X$. After replacing $Y'$ by an étale covering by a scheme (which we are allowed by our preliminary remark), this follows from Morphisms, Lemma 29.58.2. $\square$
Your email address will not be published.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)