Lemma 66.19.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $\overline{x} : \mathop{\mathrm{Spec}}(k) \to X$ be a geometric point of $X$ lying over $x \in |X|$. Let $\varphi : U \to X$ be an étale morphism of algebraic spaces and let $u \in |U|$ with $\varphi (u) = x$. Then there exists a geometric point $\overline{u} : \mathop{\mathrm{Spec}}(k) \to U$ lying over $u$ with $\overline{x} = \varphi \circ \overline{u}$.
Proof. Choose an affine scheme $U'$ with $u' \in U'$ and an étale morphism $U' \to U$ which maps $u'$ to $u$. If we can prove the lemma for $(U', u') \to (X, x)$ then the lemma follows. Hence we may assume that $U$ is a scheme, in particular that $U \to X$ is representable. Then look at the cartesian diagram
\[ \xymatrix{ \mathop{\mathrm{Spec}}(k) \times _{\overline{x}, X, \varphi } U \ar[d]_{\text{pr}_1} \ar[r]_-{\text{pr}_2} & U \ar[d]^\varphi \\ \mathop{\mathrm{Spec}}(k) \ar[r]^-{\overline{x}} & X } \]
The projection $\text{pr}_1$ is the base change of an étale morphisms so it is étale, see Lemma 66.16.5. Therefore, the scheme $\mathop{\mathrm{Spec}}(k) \times _{\overline{x}, X, \varphi } U$ is a disjoint union of finite separable extensions of $k$, see Morphisms, Lemma 29.36.7. But $k$ is algebraically closed, so all these extensions are trivial, so $\mathop{\mathrm{Spec}}(k) \times _{\overline{x}, X, \varphi } U$ is a disjoint union of copies of $\mathop{\mathrm{Spec}}(k)$ and each of these corresponds to a geometric point $\overline{u}$ with $\varphi \circ \overline{u} = \overline{x}$. By Lemma 66.4.3 the map
\[ |\mathop{\mathrm{Spec}}(k) \times _{\overline{x}, X, \varphi } U| \longrightarrow |\mathop{\mathrm{Spec}}(k)| \times _{|X|} |U| \]
is surjective, hence we can pick $\overline{u}$ to lie over $u$. $\square$
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