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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