Lemma 68.11.6. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $x$ be a point of $X$. The category of elementary étale neighborhoods of $(X, x)$ is cofiltered (see Categories, Definition 4.20.1).
Proof. The category is nonempty by Lemma 68.11.4. Suppose that we have two elementary étale neighbourhoods $(U_ i, u_ i) \to (X, x)$. Then consider $U = U_1 \times _ X U_2$. Since $\mathop{\mathrm{Spec}}(\kappa (u_ i)) \to X$, $i = 1, 2$ are both monomorphisms in the class of $x$ (Lemma 68.11.3) , we see that
\[ u = \mathop{\mathrm{Spec}}(\kappa (u_1)) \times _ X \mathop{\mathrm{Spec}}(\kappa (u_2)) \]
is the spectrum of a field $\kappa (u)$ such that the induced maps $\kappa (u_ i) \to \kappa (u)$ are isomorphisms. Then $u \to U$ is a point of $U$ and we see that $(U, u) \to (X, x)$ is an elementary étale neighbourhood dominating $(U_ i, u_ i)$. If $a, b : (U_1, u_1) \to (U_2, u_2)$ are two morphisms between our elementary étale neighbourhoods, then we consider the scheme
\[ U = U_1 \times _{(a, b), (U_2 \times _ X U_2), \Delta } U_2 \]
Using Properties of Spaces, Lemma 66.16.6 we see that $U \to X$ is étale. Moreover, in exactly the same manner as before we see that $U$ has a point $u$ such that $(U, u) \to (X, x)$ is an elementary étale neighbourhood. Finally, $U \to U_1$ equalizes $a$ and $b$ and the proof is finished. $\square$
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