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The Stacks project

Lemma 59.29.4. Let S be a scheme, and let \overline{s} be a geometric point of S. The category of étale neighborhoods is cofiltered. More precisely:

  1. Let (U_ i, \overline{u}_ i)_{i = 1, 2} be two étale neighborhoods of \overline{s} in S. Then there exists a third étale neighborhood (U, \overline{u}) and morphisms (U, \overline{u}) \to (U_ i, \overline{u}_ i), i = 1, 2.

  2. Let h_1, h_2: (U, \overline{u}) \to (U', \overline{u}') be two morphisms between étale neighborhoods of \overline{s}. Then there exist an étale neighborhood (U'', \overline{u}'') and a morphism h : (U'', \overline{u}'') \to (U, \overline{u}) which equalizes h_1 and h_2, i.e., such that h_1 \circ h = h_2 \circ h.

Proof. For part (1), consider the fibre product U = U_1 \times _ S U_2. It is étale over both U_1 and U_2 because étale morphisms are preserved under base change, see Proposition 59.26.2. The map \overline{s} \to U defined by (\overline{u}_1, \overline{u}_2) gives it the structure of an étale neighborhood mapping to both U_1 and U_2. For part (2), define U'' as the fibre product

\xymatrix{ U'' \ar[r] \ar[d] & U \ar[d]^{(h_1, h_2)} \\ U' \ar[r]^-\Delta & U' \times _ S U'. }

Since \overline{u} and \overline{u}' agree over S with \overline{s}, we see that \overline{u}'' = (\overline{u}, \overline{u}') is a geometric point of U''. In particular U'' \not= \emptyset . Moreover, since U' is étale over S, so is the fibre product U'\times _ S U' (see Proposition 59.26.2). Hence the vertical arrow (h_1, h_2) is étale by Remark 59.29.2 above. Therefore U'' is étale over U' by base change, and hence also étale over S (because compositions of étale morphisms are étale). Thus (U'', \overline{u}'') is a solution to the problem. \square


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