**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 (2) 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 60.4.4 or Lemma 60.4.5) and continuous by definition of the topology on $|X|$. The top horizontal arrow is surjective, continuous, and open by Morphisms, Lemma 28.34.13. The left vertical arrow is continuous and open (by Morphisms, Lemma 28.34.13 again.) Hence it follows formally that the right vertical arrow is continuous and open.

To finish the proof we prove (1). Let $a : X \to Y$ be a morphism of algebraic spaces. According to Spaces, Lemma 59.11.6 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.
$\square$

## Comments (3)

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