Lemma 66.19.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $\overline{x}$ be a geometric point of $X$. The category of étale neighborhoods is cofiltered. More precisely:
Let $(U_ i, \overline{u}_ i)_{i = 1, 2}$ be two étale neighborhoods of $\overline{x}$ in $X$. Then there exists a third étale neighborhood $(U, \overline{u})$ and morphisms $(U, \overline{u}) \to (U_ i, \overline{u}_ i)$, $i = 1, 2$.
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 $h_1 \circ h = h_2 \circ h$.
Moreover, given any étale neighbourhood $(U, \overline{u}) \to (X, \overline{x})$ there exists a morphism of étale neighbourhoods $(U', \overline{u}') \to (U, \overline{u})$ where $U'$ is a scheme.
Proof.
For part (1), consider the fibre product $U = U_1 \times _ X U_2$. It is étale over both $U_1$ and $U_2$ because étale morphisms are preserved under base change and composition, see Lemmas 66.16.5 and 66.16.4. The map $\overline{u} \to U$ 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
\[ \xymatrix{ U'' \ar[r] \ar[d] & U \ar[d]^{(h_1, h_2)} \\ U' \ar[r]^-\Delta & U' \times _ X U'. } \]
Since $\overline{u}$ and $\overline{u}'$ agree over $X$ with $\overline{x}$, 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 $X$, so is the fibre product $U'\times _ X U'$ (as seen above in the case of $U_1 \times _ X U_2$). Hence the vertical arrow $(h_1, h_2)$ is étale by Lemma 66.16.6. Therefore $U''$ is étale over $U'$ by base change, and hence also étale over $X$ (because compositions of étale morphisms are étale). Thus $(U'', \overline{u}'')$ is a solution to the problem posed by (2).
To see the final assertion, choose any surjective étale morphism $U' \to U$ where $U'$ is a scheme. Then $U' \times _ U \overline{u}$ is a scheme surjective and étale over $\overline{u} = \mathop{\mathrm{Spec}}(k)$ with $k$ algebraically closed. It follows (see Morphisms, Lemma 29.36.7) that $U' \times _ U \overline{u} \to \overline{u}$ has a section which gives us the desired $\overline{u}'$.
$\square$
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