The Stacks project

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

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 $\overline{u}'$. $\square$

Proof. Choose an affine scheme $U'$ with $u' \in U'$ and an étale morphism $U' \to U$ which maps $u'$ to $u$. If we can prove the lemma for $(U', u') \to (X, x)$ then the lemma follows. Hence we may assume that $U$ is a scheme, in particular that $U \to X$ is representable. Then look at the cartesian diagram

The projection $\text{pr}_1$ is the base change of an étale morphisms so it is étale, see Lemma 66.16.5. Therefore, the scheme $\mathop{\mathrm{Spec}}(k) \times _{\overline{x}, X, \varphi } U$ is a disjoint union of finite separable extensions of $k$, see Morphisms, Lemma 29.36.7. But $k$ is algebraically closed, so all these extensions are trivial, so $\mathop{\mathrm{Spec}}(k) \times _{\overline{x}, X, \varphi } U$ is a disjoint union of copies of $\mathop{\mathrm{Spec}}(k)$ and each of these corresponds to a geometric point $\overline{u}$ with $\varphi \circ \overline{u} = \overline{x}$. By Lemma 66.4.3 the map

is surjective, hence we can pick $\overline{u}$ to lie over $u$. $\square$

Proof. Let $u \in |U|$ be the image of $\overline{u}$. As $|U| = \bigcup _{i \in I} \varphi _ i(|U_ i|)$ there exists an $i$ and a point $u_ i \in U_ i$ mapping to $x$. Apply Lemma 66.19.4 to $(U_ i, u_ i) \to (U, u)$ and $\overline{u}$ to get the desired geometric point. $\square$

Proof. In the proof of Lemma 66.19.5 we have seen that the scheme $U_{\overline{x}}$ is a disjoint union of schemes isomorphic to $\overline{x}$. Thus we can also think of $|U_{\overline{x}}|$ as the set of geometric points of $U$ lying over $\overline{x}$, i.e., as the collection of morphisms $\overline{u} : \overline{x} \to U$ fitting into the diagram of Definition 66.19.2. From this it follows that $u(X)$ is a singleton, and that $u(U \times _ V W) = u(U) \times _{u(V)} u(W)$ whenever $U \to V$ and $W \to V$ are morphisms in $X_{\acute{e}tale}$. And, given a covering $\{ U_ i \to U\} _{i \in I}$ in $X_{\acute{e}tale}$ we see that $\coprod u(U_ i) \to u(U)$ is surjective by Lemma 66.19.5. Hence Sites, Proposition 7.33.3 applies, so $p$ is a point of the site $X_{\acute{e}tale}$. Finally, the our functor $\mathcal{F} \mapsto \mathcal{F}_{\overline{s}}$ is given by exactly the same colimit as the functor $\mathcal{F} \mapsto \mathcal{F}_ p$ associated to $p$ in Sites, Equation 7.32.1.1 which proves the final assertion. $\square$

Proof. This result follows from the general material in Modules on Sites, Section 18.36. This is true because $\mathcal{F} \mapsto \mathcal{F}_{\overline{x}}$ comes from a point of the small étale site of $X$, see Lemma 66.19.7. See the proof of Étale Cohomology, Lemma 59.29.9 for a direct proof of some of these statements in the setting of the small étale site of a scheme. $\square$

Proof. Recall that $f_{small}$ is defined via $f_{spaces, small}$ in Lemma 66.18.8. Parts (1), (2) and (3) are general consequences of the fact that $f_{spaces, {\acute{e}tale}} : X_{spaces, {\acute{e}tale}} \to Y_{spaces, {\acute{e}tale}}$ is a morphism of sites, see Sites, Definition 7.14.1 for (2), Modules on Sites, Lemma 18.31.2 for (1), and Sites, Lemma 7.13.5 for (3).

Proof of (4). This statement is a special case of Sites, Lemma 7.34.2 via Lemma 66.19.7. We also provide a direct proof. Note that by Lemma 66.19.8. taking stalks commutes with sheafification. Let $\mathcal{G}'$ be the sheaf on $Y_{spaces, {\acute{e}tale}}$ whose restriction to $Y_{\acute{e}tale}$ is $\mathcal{G}$. Recall that $f_{spaces, {\acute{e}tale}}^{-1}\mathcal{G}'$ is the sheaf associated to the presheaf

see Sites, Sections 7.13 and 7.5. Thus we have

in the third equality the pair $(U, \overline{u})$ and the map $a : U \to X \times _ Y V$ corresponds to the pair $(V, a \circ \overline{u})$. Since the stalk of $\mathcal{G}'$ (resp. $f_{spaces, {\acute{e}tale}}^{-1}\mathcal{G}'$) agrees with the stalk of $\mathcal{G}$ (resp. $f_{small}^{-1}\mathcal{G}$), see Equation (66.19.6.1) the result follows. $\square$

Proof. We know the theorem is true if $X$ is a scheme, see Étale Cohomology, Theorem 59.29.10. Choose a surjective étale morphism $f : U \to X$ where $U$ is a scheme. Since $\{ U \to X\} $ is a covering (in $X_{spaces, {\acute{e}tale}}$) we can check whether a map of sheaves is injective, or surjective by restricting to $U$. Now if $\overline{u} : \mathop{\mathrm{Spec}}(k) \to U$ is a geometric point of $U$, then $(\mathcal{F}|_ U)_{\overline{u}} = \mathcal{F}_{\overline{x}}$ where $\overline{x} = f \circ \overline{u}$. (This is clear from the colimits defining the stalks at $\overline{u}$ and $\overline{x}$, but it also follows from Lemma 66.19.9.) Hence the result for $U$ implies the result for $X$ and we win. $\square$

Proof. By Sites, Lemma 7.32.7 there is a one to one correspondence between points of the site and points of the associated topos. Hence we may assume that $p$ is given by a functor $u : X_{\acute{e}tale}\to \textit{Sets}$ which defines a point of the site $X_{\acute{e}tale}$. Let $U \in \mathop{\mathrm{Ob}}\nolimits (X_{\acute{e}tale})$ be an object whose structure morphism $j : U \to X$ is surjective. Note that $h_ U$ is a sheaf which surjects onto the final sheaf. Since taking stalks is exact we see that $(h_ U)_ p = u(U)$ is not empty (use Sites, Lemma 7.32.3). Pick $x \in u(U)$. By Sites, Lemma 7.35.1 we obtain a point $q : \mathop{\mathit{Sh}}\nolimits (pt) \to \mathop{\mathit{Sh}}\nolimits (U_{\acute{e}tale})$ such that $p = j_{small} \circ q$, so that $\mathcal{F}_ p = (\mathcal{F}|_ U)_ q$ functorially. By Étale Cohomology, Lemma 59.29.12 there is a geometric point $\overline{u}$ of $U$ and a functorial isomorphism $\mathcal{G}_ q = \mathcal{G}_{\overline{u}}$ for $\mathcal{G} \in \mathop{\mathit{Sh}}\nolimits (U_{\acute{e}tale})$. Set $\overline{x} = j \circ \overline{u}$. Then we see that $\mathcal{F}_{\overline{x}} \cong (\mathcal{F}|_ U)_{\overline{u}}$ functorially in $\mathcal{F}$ on $X_{\acute{e}tale}$ by Lemma 66.19.9 and we win. $\square$