Lemma 59.29.5. Let $S$ be a scheme. Let $\overline{s}$ be a geometric point of $S$. Let $(U, \overline{u})$ be an étale neighborhood of $\overline{s}$. Let $\mathcal{U} = \{ \varphi _ i : U_ i \to U \} _{i\in I}$ be an étale covering. Then there exist $i \in I$ and $\overline{u}_ i : \overline{s} \to U_ i$ such that $\varphi _ i : (U_ i, \overline{u}_ i) \to (U, \overline{u})$ is a morphism of étale neighborhoods.

**Proof.**
As $U = \bigcup _{i\in I} \varphi _ i(U_ i)$, the fibre product $\overline{s} \times _{\overline{u}, U, \varphi _ i} U_ i$ is not empty for some $i$. Then look at the cartesian diagram

The projection $\text{pr}_1$ is the base change of an étale morphisms so it is étale, see Proposition 59.26.2. Therefore, $\overline{s} \times _{\overline{u}, U, \varphi _ i} U_ i$ is a disjoint union of finite separable extensions of $k$, by Proposition 59.26.2. Here $\overline{s} = \mathop{\mathrm{Spec}}(k)$. But $k$ is algebraically closed, so all these extensions are trivial, and there exists a section $\sigma $ of $\text{pr}_1$. The composition $\text{pr}_2 \circ \sigma $ gives a map compatible with $\overline{u}$. $\square$

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