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
\xymatrix{ \overline{s} \times _{\overline{u}, U, \varphi _ i} U_ i \ar[d]^{\text{pr}_1} \ar[r]_-{\text{pr}_2} & U_ i \ar[d]^{\varphi _ i} \\ \mathop{\mathrm{Spec}}(k) = \overline{s} \ar@/^1pc/[u]^\sigma \ar[r]^-{\overline{u}} & U }
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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