Lemma 67.13.9. 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 67.7.2. The morphism $Y' \to U$ maps $|Y'|$ into $E$. By our construction of $Z_ x$ in the proof of Lemma 67.13.5 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 29.57.2.
$\square$

## Comments (0)