The Stacks project

Lemma 68.7.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x, x' \in |X|$ and assume $x' \leadsto x$, i.e., $x$ is a specialization of $x'$. Assume the pair $(X, x')$ satisfies the equivalent conditions of Lemma 68.4.5. Then for every ├ętale morphism $\varphi : U \to X$ from a scheme $U$ and any $u \in U$ with $\varphi (u) = x$, exists a point $u'\in U$, $u' \leadsto u$ with $\varphi (u') = x'$.

Proof. We may replace $U$ by an affine open neighbourhood of $u$. Hence we may assume that $U$ is affine. As $x$ is in the image of the open map $|U| \to |X|$, so is $x'$. Thus we may replace $X$ by the Zariski open subspace corresponding to the image of $|U| \to |X|$, see Properties of Spaces, Lemma 66.4.10. In other words we may assume that $U \to X$ is surjective and ├ętale. Let $s, t : R = U \times _ X U \to U$ be the projections. By our assumption that $(X, x')$ satisfies the equivalent conditions of Lemma 68.4.5 we see that the fibres of $|U| \to |X|$ and $|R| \to |X|$ over $x'$ are finite. Say $\{ u'_1, \ldots , u'_ n\} \subset U$ and $\{ r'_1, \ldots , r'_ m\} \subset R$ form the complete inverse image of $\{ x'\} $. Consider the closed sets

\[ T = \overline{\{ u'_1\} } \cup \ldots \cup \overline{\{ u'_ n\} } \subset |U|, \quad T' = \overline{\{ r'_1\} } \cup \ldots \cup \overline{\{ r'_ m\} } \subset |R|. \]

Trivially we have $s(T') \subset T$. Because $R$ is an equivalence relation we also have $t(T') = s(T')$ as the set $\{ r_ j'\} $ is invariant under the inverse of $R$ by construction. Let $w \in T$ be any point. Then $u'_ i \leadsto w$ for some $i$. Choose $r \in R$ with $s(r) = w$. Since generalizations lift along $s : R \to U$, see Remark 68.4.1, we can find $r' \leadsto r$ with $s(r') = u_ i'$. Then $r' = r'_ j$ for some $j$ and we conclude that $w \in s(T')$. Hence $T = s(T') = t(T')$ is an $|R|$-invariant closed set in $|U|$. This means $T$ is the inverse image of a closed (!) subset $T'' = \varphi (T)$ of $|X|$, see Properties of Spaces, Lemmas 66.4.5 and 66.4.6. Hence $T'' = \overline{\{ x'\} }$. Thus $T$ contains some point $u_1$ mapping to $x$ as $x \in T''$. I.e., we see that for some $i$ there exists a specialization $u'_ i \leadsto u_1$ which maps to the given specialization $x' \leadsto x$.

To finish the proof, choose a point $r \in R$ such that $s(r) = u$ and $t(r) = u_1$ (using Properties of Spaces, Lemma 66.4.3). As generalizations lift along $t$, and $u'_ i \leadsto u_1$ we can find a specialization $r' \leadsto r$ such that $t(r') = u'_ i$. Set $u' = s(r')$. Then $u' \leadsto u$ and $\varphi (u') = x'$ as desired. $\square$

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