Lemma 66.4.6. Let $S$ be a scheme. There exists a unique topology on the sets of points of algebraic spaces over $S$ with the following properties:

1. if $X$ is a scheme over $S$, then the topology on $|X|$ is the usual one (via the identification of Lemma 66.4.2),

2. for every morphism of algebraic spaces $X \to Y$ over $S$ the map $|X| \to |Y|$ is continuous, and

3. for every étale morphism $U \to X$ with $U$ a scheme the map of topological spaces $|U| \to |X|$ is continuous and open.

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

$\xymatrix{ U'' \ar[r] \ar[d] & U' \ar[d] \\ U \ar[r] & X }$

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

$\xymatrix{ |U''| \ar[r] \ar[d] & |U'| \ar[d] \\ |U| \ar[r] & |X| }$

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

$\xymatrix{ U \ar[d]_ p \ar[r]_\alpha & V \ar[d]^ q \\ X \ar[r]^ a & Y }$

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

$\xymatrix{ |U| \ar[d]_ p \ar[r]_\alpha & |V| \ar[d]^ q \\ |X| \ar[r]^ a & |Y| }$

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. $\square$

Comment #2333 by Wessel on

I'm probably being too pedantic, but shouldn't the lemma include a phrase along the lines of "if X is a scheme, then |X| is endowed with the Zariski topology"? Otherwise, technically, we could endow everything with the discrete topology. A similar remark applies to Tag 04XL.

Comment #2404 by on

OK, it is not pedantic as much as cautious... technically you are correct... but I am going to leave it as is... for now...

Comment #4214 by David Holmes on

I came here to make the same comment as Wessel, but then I saw Wessel's comment, so I won't (maybe I just did...). So maybe count this as another vote for changing it??

Comment #4396 by on

OK, I have added it as a condition in the lemma. But I am leaving the case of Lemma 100.4.7 alone for now because the typography suggests that we are dealing with the space associated to $\mathcal{X}$ or $X$ in different ways (for one we are trying to define the topology and for the second we already have it). See changes here.

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• 4 comment(s) on Section 66.4: Points of algebraic spaces

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