The Stacks project

Proof. This is Schemes, Lemma 26.13.3. $\square$

Proof. Namely, suppose given fields $K$, $L$ and morphisms $\mathop{\mathrm{Spec}}(K) \to X$, $\mathop{\mathrm{Spec}}(L) \to Z$, then the assumption that they agree as elements of $|Y|$ means that there is a common extension $M/K$ and $M/L$ such that $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(K) \to X \to Y$ and $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(L) \to Z \to Y$ agree. And this is exactly the condition which says you get a morphism $\mathop{\mathrm{Spec}}(M) \to Z \times _ Y X$. $\square$

Proof. Assume (1). Let $x : \mathop{\mathrm{Spec}}(K) \to X$ be a morphism from the spectrum of a field into $X$. By assumption the morphism of schemes $\mathop{\mathrm{Spec}}(K) \times _ X T \to \mathop{\mathrm{Spec}}(K)$ is surjective. Hence there exists a field extension $K'/K$ and a morphism $\mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K) \times _ X T$ such that the left square in the diagram

is commutative. This shows that $|f| : |T| \to |X|$ is surjective.

Assume (2). Let $Z \to X$ be a morphism where $Z$ is a scheme. We have to show that the morphism of schemes $Z \times _ X T \to T$ is surjective, i.e., that $|Z \times _ X T| \to |Z|$ is surjective. This follows from (2) and Lemma 66.4.3. $\square$

Proof. The assumption means that $U$ is a scheme, $p : U \to X$ is a surjective, étale morphism, $R = U \times _ X U$ is a scheme and defines an étale equivalence relation on $U$ such that $X = U/R$ as sheaves. By Lemma 66.4.4 we see that $|U| \to |X|$ is surjective. By Lemma 66.4.3 the map

is surjective. Hence the image of $|R| \to |U| \times |U|$ is exactly the set of pairs $(u_1, u_2) \in |U| \times |U|$ such that $u_1$ and $u_2$ have the same image in $|X|$. Combining these two statements we get the result of the lemma. $\square$

Proof. Let $X$ be an algebraic space over $S$. Let $p : U \to X$ be a surjective étale morphism where $U$ is a scheme over $S$. We define $W \subset |X|$ is open if and only if $|p|^{-1}(W)$ is an open subset of $|U|$. This is a topology on $|X|$ (it is the quotient topology on $|X|$, see Topology, Lemma 5.6.2).

Let us prove that the topology is independent of the choice of the presentation. To do this it suffices to show that if $U'$ is a scheme, and $U' \to X$ is an étale morphism, then the map $|U'| \to |X|$ (with topology on $|X|$ defined using $U \to X$ as above) is open and continuous; which in addition will prove that (3) holds. Set $U'' = U \times _ X U'$, so that we have the commutative diagram

As $U \to X$ and $U' \to X$ are étale we see that both $U'' \to U$ and $U'' \to U'$ are étale morphisms of schemes. Moreover, $U'' \to U'$ is surjective. Hence we get a commutative diagram of maps of sets

The lower horizontal arrow is surjective (see Lemma 66.4.4 or Lemma 66.4.5) and continuous by definition of the topology on $|X|$. The top horizontal arrow is surjective, continuous, and open by Morphisms, Lemma 29.36.13. The left vertical arrow is continuous and open (by Morphisms, Lemma 29.36.13 again.) Hence it follows formally that the right vertical arrow is continuous and open.

To finish the proof we prove (2). Let $a : X \to Y$ be a morphism of algebraic spaces. According to Spaces, Lemma 65.11.6 we can find a diagram

where $U$ and $V$ are schemes, and $p$ and $q$ are surjective and étale. This gives rise to the diagram

where all but the lower horizontal arrows are known to be continuous and the two vertical arrows are surjective and open. It follows that the lower horizontal arrow is continuous as desired. $\square$

Proof. In order to prove (1) let us construct the inverse of the rule. Namely, suppose that $W \subset |X|$ is open. Choose a presentation $X = U/R$ corresponding to the surjective étale map $p : U \to X$ and étale maps $s, t : R \to U$. By construction we see that $|p|^{-1}(W)$ is an open of $U$. Denote $W' \subset U$ the corresponding open subscheme. It is clear that $R' = s^{-1}(W') = t^{-1}(W')$ is a Zariski open of $R$ which defines an étale equivalence relation on $W'$. By Spaces, Lemma 65.10.2 the morphism $X' = W'/R' \to X$ is an open immersion. Hence $X'$ is an algebraic space by Spaces, Lemma 65.11.3. By construction $|X'| = W$, i.e., $X'$ is a subspace of $X$ corresponding to $W$. Thus (1) is proved.

To prove (2), note that if $\{ X_ i \subset X\} _{i \in I}$ is a collection of open subspaces, then it is a Zariski covering if and only if the $U = \bigcup U \times _ X X_ i$ is an open covering. This follows from the definition of a Zariski covering and the fact that the morphism $U \to X$ is surjective as a map of presheaves on $(\mathit{Sch}/S)_{fppf}$. On the other hand, we see that $|X| = \bigcup |X_ i|$ if and only if $U = \bigcup U \times _ X X_ i$ by Lemma 66.4.5 (and the fact that the projections $U \times _ X X_ i \to X_ i$ are surjective and étale). Thus the equivalence of (2) follows. $\square$

Proof. By Spaces, Lemma 65.12.3 we see that $Y' = Y \times _ X X' \to Y$ is an open immersion. If $|f|(|Y|) \subset |X'|$, then clearly $|Y'| = |Y|$. Hence $Y' = Y$ by Lemma 66.4.8. $\square$

Proof. The existence of the factorization follows from Lemma 66.4.9. The morphism $f'$ is surjective according to Lemma 66.4.4. To see $f'$ is étale, suppose that $T \to X'$ is a morphism where $T$ is a scheme. Then $T \times _ X U = T \times _{X'} U$ as $X' \to X$ is a monomorphism of sheaves. Thus the projection $T \times _{X'} U \to T$ is étale as we assumed $f$ étale. We have $U \times _ X U = U \times _{X'} U$ as $X' \to X$ is a monomorphism. Then $X' = U/R$ follows from Spaces, Lemma 65.9.1. $\square$

Proof. Since $p$ and $q$ define the same point of $|X|$, we see that the scheme

is nonempty. Since the base change of a monomorphism is a monomorphism this means that the projection morphism $Y \to \mathop{\mathrm{Spec}}(L)$ is a monomorphism. Hence $Y = \mathop{\mathrm{Spec}}(L)$, see Schemes, Lemma 26.23.11. We conclude that $q$ factors through $p$. Uniqueness comes from the fact that $p$ is a monomorphism. $\square$

Proof. This follows from Lemma 66.4.11. $\square$