The Stacks project

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

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. $\square$

Proof. As $U = \bigcup _{i\in I} \varphi _ i(U_ i)$, the fibre product $\overline{s} \times _{\overline{u}, U, \varphi _ i} U_ i$ is not empty for some $i$. Then look at the cartesian diagram

The projection $\text{pr}_1$ is the base change of an étale morphisms so it is étale, see Proposition 59.26.2. Therefore, $\overline{s} \times _{\overline{u}, U, \varphi _ i} U_ i$ is a disjoint union of finite separable extensions of $k$, by Proposition 59.26.2. Here $\overline{s} = \mathop{\mathrm{Spec}}(k)$. But $k$ is algebraically closed, so all these extensions are trivial, and there exists a section $\sigma $ of $\text{pr}_1$. The composition $\text{pr}_2 \circ \sigma $ gives a map compatible with $\overline{u}$. $\square$

Proof. In the proof of Lemma 59.29.5 we have seen that the scheme $U_{\overline{s}}$ is a disjoint union of schemes isomorphic to $\overline{s}$. Thus we can also think of $|U_{\overline{s}}|$ as the set of geometric points of $U$ lying over $\overline{s}$, i.e., as the collection of morphisms $\overline{u} : \overline{s} \to U$ fitting into the diagram of Definition 59.29.1. From this it follows that $u(S)$ 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 $S_{\acute{e}tale}$. And, given a covering $\{ U_ i \to U\} _{i \in I}$ in $S_{\acute{e}tale}$ we see that $\coprod u(U_ i) \to u(U)$ is surjective by Lemma 59.29.5. Hence Sites, Proposition 7.33.3 applies, so $p$ is a point of the site $S_{\acute{e}tale}$. Finally, 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. Before we indicate how to prove this by direct arguments we note that the result follows from the general material in Modules on Sites, Section 18.36. This is true because $\mathcal{F} \mapsto \mathcal{F}_{\overline{s}}$ comes from a point of the small étale site of $S$, see Lemma 59.29.7. We will only give a direct proof of (1), (2) and (3), and omit a direct proof of (4).

Exactness as a functor on $\textit{PAb}(S_{\acute{e}tale})$ is formal from the fact that directed colimits commute with all colimits and with finite limits. The identification of the stalks in (2) is via the map

induced by the natural morphism $\mathcal{F}\to \mathcal{F}^\# $, see Theorem 59.13.2. We claim that this map is an isomorphism of abelian groups. We will show injectivity and omit the proof of surjectivity.

Let $\sigma \in \mathcal{F}_{\overline{s}}$. There exists an étale neighborhood $(U, \overline{u})\to (S, \overline{s})$ such that $\sigma $ is the image of some section $s \in \mathcal{F}(U)$. If $\kappa (\sigma ) = 0$ in $(\mathcal{F}^\# )_{\overline{s}}$ then there exists a morphism of étale neighborhoods $(U', \overline{u}')\to (U, \overline{u})$ such that $s|_{U'}$ is zero in $\mathcal{F}^\# (U')$. It follows there exists an étale covering $\{ U_ i'\to U'\} _{i\in I}$ such that $s|_{U_ i'}=0$ in $\mathcal{F}(U_ i')$ for all $i$. By Lemma 59.29.5 there exist $i \in I$ and a morphism $\overline{u}_ i': \overline{s} \to U_ i'$ such that $(U_ i', \overline{u}_ i') \to (U', \overline{u}')\to (U, \overline{u})$ are morphisms of étale neighborhoods. Hence $\sigma = 0$ since $(U_ i', \overline{u}_ i') \to (U, \overline{u})$ is a morphism of étale neighbourhoods such that we have $s|_{U'_ i}=0$. This proves $\kappa $ is injective.

To show that the functor $\textit{Ab}(S_{\acute{e}tale}) \to \textit{Ab}$ is exact, consider any short exact sequence in $\textit{Ab}(S_{\acute{e}tale})$: $ 0\to \mathcal{F}\to \mathcal{G}\to \mathcal H \to 0. $ This gives us the exact sequence of presheaves

where $/^ p$ denotes the quotient in $\textit{PAb}(S_{\acute{e}tale})$. Taking stalks at $\overline{s}$, we see that $(\mathcal H /^ p\mathcal{G})_{\bar{s}} = (\mathcal H /\mathcal{G})_{\bar{s}} = 0$, since the sheafification of $\mathcal H/^ p\mathcal{G}$ is $0$. Therefore,

is exact, since taking stalks is exact as a functor from presheaves. $\square$

Proof. The necessity of exactness on stalks follows from Lemma 59.29.9. For the converse, it suffices to show that a map of sheaves is surjective (respectively injective) if and only if it is surjective (respectively injective) on all stalks. We prove this in the case of surjectivity, and omit the proof in the case of injectivity.

Let $\alpha : \mathcal{F} \to \mathcal{G}$ be a map of sheaves such that $\mathcal{F}_{\overline{s}} \to \mathcal{G}_{\overline{s}}$ is surjective for all geometric points. Fix $U \in \mathop{\mathrm{Ob}}\nolimits (S_{\acute{e}tale})$ and $s \in \mathcal{G}(U)$. For every $u \in U$ choose some $\overline{u} \to U$ lying over $u$ and an étale neighborhood $(V_ u , \overline{v}_ u) \to (U, \overline{u})$ such that $s|_{V_ u} = \alpha (s_{V_ u})$ for some $s_{V_ u} \in \mathcal{F}(V_ u)$. This is possible since $\alpha $ is surjective on stalks. Then $\{ V_ u \to U\} _{u \in U}$ is an étale covering on which the restrictions of $s$ are in the image of the map $\alpha $. Thus, $\alpha $ is surjective, see Sites, Section 7.11. $\square$

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 it suffices to prove (1). By Sites, Proposition 7.33.3 the functor $u$ has the following properties: (a) $u(S) = \{ *\} $, (b) $u(U \times _ V W) = u(U) \times _{u(V)} u(W)$, and (c) if $\{ U_ i \to U\} $ is an étale covering, then $\coprod u(U_ i) \to u(U)$ is surjective. In particular, if $U' \subset U$ is an open subscheme, then $u(U') \subset u(U)$. Moreover, by Sites, Lemma 7.32.7 we can write $u(U) = p^{-1}(h_ U^\# )$, in other words $u(U)$ is the stalk of the representable sheaf $h_ U$. If $U = V \amalg W$, then we see that $h_ U = (h_ V \amalg h_ W)^\# $ and we get $u(U) = u(V) \amalg u(W)$ since $p^{-1}$ is exact.

Consider the restriction of $u$ to $S_{Zar}$. By Sites, Examples 7.33.5 and 7.33.6 there exists a unique point $s \in S$ such that for $S' \subset S$ open we have $u(S') = \{ *\} $ if $s \in S'$ and $u(S') = \emptyset $ if $s \not\in S'$. Note that if $\varphi : U \to S$ is an object of $S_{\acute{e}tale}$ then $\varphi (U) \subset S$ is open (see Proposition 59.26.2) and $\{ U \to \varphi (U)\} $ is an étale covering. Hence we conclude that $u(U) = \emptyset \Leftrightarrow s \in \varphi (U)$.

Pick a geometric point $\overline{s} : \overline{s} \to S$ lying over $s$, see Definition 59.29.1 for customary abuse of notation. Suppose that $\varphi : U \to S$ is an object of $S_{\acute{e}tale}$ with $U$ affine. Note that $\varphi $ is separated, and that the fibre $U_ s$ of $\varphi $ over $s$ is an affine scheme over $\mathop{\mathrm{Spec}}(\kappa (s))$ which is the spectrum of a finite product of finite separable extensions $k_ i$ of $\kappa (s)$. Hence we may apply Étale Morphisms, Lemma 41.18.2 to get an étale neighbourhood $(V, \overline{v})$ of $(S, \overline{s})$ such that

with $U_ i \to V$ an isomorphism and $W$ having no point lying over $\overline{v}$. Thus we conclude that

and of course also $u(U_ i) = u(V)$. After shrinking $V$ a bit we can assume that $V$ has exactly one point lying over $s$, and hence $W$ has no point lying over $s$. By the above this then gives $u(W) = \emptyset $. Hence we obtain

Note that $u(V) \not= \emptyset $ as $s$ is in the image of $V \to S$. In particular, we see that in this situation $u(U)$ is a finite set with $n$ elements.

Consider the limit

over the category of étale neighbourhoods $(V, \overline{v})$ of $\overline{s}$. It is clear that we get the same value when taking the limit over the subcategory of $(V, \overline{v})$ with $V$ affine. By the previous paragraph (applied with the roles of $V$ and $U$ switched) we see that in this case $u(V)$ is always a finite nonempty set. Moreover, the limit is cofiltered, see Lemma 59.29.4. Hence by Categories, Section 4.20 the limit is nonempty. Pick an element $x$ from this limit. This means we obtain a $x_{V, \overline{v}} \in u(V)$ for every étale neighbourhood $(V, \overline{v})$ of $(S, \overline{s})$ such that for every morphism of étale neighbourhoods $\varphi : (V', \overline{v}') \to (V, \overline{v})$ we have $u(\varphi )(x_{V', \overline{v}'}) = x_{V, \overline{v}}$.

We will use the choice of $x$ to construct a functorial bijective map

for $U \in \mathop{\mathrm{Ob}}\nolimits (S_{\acute{e}tale})$ which will conclude the proof. See Lemma 59.29.7 and its proof for a description of $|U_{\overline{s}}|$. First we claim that it suffices to construct the map for $U$ affine. We omit the proof of this claim. Assume $U \to S$ in $S_{\acute{e}tale}$ with $U$ affine, and let $\overline{u} : \overline{s} \to U$ be an element of $|U_{\overline{s}}|$. Choose a $(V, \overline{v})$ such that $U \times _ S V$ decomposes as in the third paragraph of the proof. Then the pair $(\overline{u}, \overline{v})$ gives a geometric point of $U \times _ S V$ lying over $\overline{v}$ and determines one of the components $U_ i$ of $U \times _ S V$. More precisely, there exists a section $\sigma : V \to U \times _ S V$ of the projection $\text{pr}_ U$ such that $(\overline{u}, \overline{v}) = \sigma \circ \overline{v}$. Set $c(\overline{u}) = u(\text{pr}_ U)(u(\sigma )(x_{V, \overline{v}})) \in u(U)$. We have to check this is independent of the choice of $(V, \overline{v})$. By Lemma 59.29.4 the category of étale neighbourhoods is cofiltered. Hence it suffice to show that given a morphism of étale neighbourhood $\varphi : (V', \overline{v}') \to (V, \overline{v})$ and a choice of a section $\sigma ' : V' \to U \times _ S V'$ of the projection such that $(\overline{u}, \overline{v'}) = \sigma ' \circ \overline{v}'$ we have $u(\sigma ')(x_{V', \overline{v}'}) = u(\sigma )(x_{V, \overline{v}})$. Consider the diagram

Now, it may not be the case that this diagram commutes. The reason is that the schemes $V'$ and $V$ may not be connected, and hence the decompositions used to construct $\sigma '$ and $\sigma $ above may not be unique. But we do know that $\sigma \circ \varphi \circ \overline{v}' = (1 \times \varphi ) \circ \sigma ' \circ \overline{v}'$ by construction. Hence, since $U \times _ S V$ is étale over $S$, there exists an open neighbourhood $V'' \subset V'$ of $\overline{v'}$ such that the diagram does commute when restricted to $V''$, see Morphisms, Lemma 29.35.17. This means we may extend the diagram above to

such that the left square and the outer rectangle commute. Since $u$ is a functor this implies that $x_{V'', \overline{v}'}$ maps to the same element in $u(U \times _ S V)$ no matter which route we take through the diagram. On the other hand, it maps to the elements $x_{V', \overline{v}'}$ and $x_{V, \overline{v}}$ in $u(V')$ and $u(V)$. This implies the desired equality $u(\sigma ')(x_{V', \overline{v}'}) = u(\sigma )(x_{V, \overline{v}})$.

In a similar manner one proves that the construction $c : |U_{\overline{s}}| \to u(U)$ is functorial in $U$; details omitted. And finally, by the results of the third paragraph it is clear that the map $c$ is bijective which ends the proof of the lemma. $\square$