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 étale 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 étale. 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
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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