Lemma 67.53.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. Then $f$ is a universal homeomorphism (in the sense of Section 67.3) if and only if for every morphism of algebraic spaces $Z \to Y$ the base change map $Z \times _ Y X \to Z$ induces a homeomorphism $|Z \times _ Y X| \to |Z|$.
67.53 Universal homeomorphisms
The class of universal homeomorphisms of schemes is closed under composition and arbitrary base change and is fppf local on the base. See Morphisms, Lemmas 29.45.3 and 29.45.2 and Descent, Lemma 35.23.9. Thus, if we apply the discussion in Section 67.3 to this notion we see that we know what it means for a representable morphism of algebraic spaces to be a universal homeomorphism.
Proof. If for every morphism of algebraic spaces $Z \to Y$ the base change map $Z \times _ Y X \to Z$ induces a homeomorphism $|Z \times _ Y X| \to |Z|$, then the same is true whenever $Z$ is a scheme, which formally implies that $f$ is a universal homeomorphism in the sense of Section 67.3. Conversely, if $f$ is a universal homeomorphism in the sense of Section 67.3 then $X \to Y$ is integral, universally injective and surjective (by Spaces, Lemma 65.5.8 and Morphisms, Lemma 29.45.5). Hence $f$ is universally closed, see Lemma 67.45.7 and universally injective and (universally) surjective, i.e., $f$ is a universal homeomorphism. $\square$
Definition 67.53.2. Let $S$ be a scheme. A morphism $f : X \to Y$ of algebraic spaces over $S$ is called a universal homeomorphism if and only if for every morphism of algebraic spaces $Z \to Y$ the base change $Z \times _ Y X \to Z$ induces a homeomorphism $|Z \times _ Y X| \to |Z|$.
This definition does not clash with the pre-existing definition for representable morphisms of algebraic spaces by our Lemma 67.53.1. For morphisms of algebraic spaces it is not the case that universal homeomorphisms are always integral.
Example 67.53.3. This is a continuation of Remark 67.19.4. Consider the algebraic space $X = \mathbf{A}^1_ k/\{ x \sim -x \mid x \not= 0\} $. There are morphisms such that the first arrow is étale surjective, the second arrow is universally injective, and the composition is the map $x \mapsto x^2$. Hence the composition is universally closed. Thus it follows that the map $X \to \mathbf{A}^1_ k$ is a universal homeomorphism, but $X \to \mathbf{A}^1_ k$ is not separated.
\[ \mathbf{A}^1_ k \longrightarrow X \longrightarrow \mathbf{A}^1_ k \]
Let $S$ be a scheme. Let $f : X \to Y$ be a universal homeomorphism of algebraic spaces over $S$. Then $f$ is universally closed, hence is quasi-compact, see Lemma 67.9.7. But $f$ need not be separated (see example above), and not even quasi-separated: an example is to take infinite dimensional affine space $\mathbf{A}^\infty = \mathop{\mathrm{Spec}}(k[x_1, x_2, \ldots ])$ modulo the equivalence relation given by flipping finitely many signs of nonzero coordinates (details omitted).
First we state the obligatory lemmas.
Lemma 67.53.4. The base change of a universal homeomorphism of algebraic spaces by any morphism of algebraic spaces is a universal homeomorphism.
Proof. This is immediate from the definition. $\square$
Lemma 67.53.5. The composition of a pair of universal homeomorphisms of algebraic spaces is a universal homeomorphism.
Proof. Omitted. $\square$
Lemma 67.53.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The canonical closed immersion $X_{red} \to X$ (see Properties of Spaces, Definition 66.12.5) is a universal homeomorphism.
Proof. Omitted. $\square$
We put the following result here as we do not currently have a better place to put it.
Lemma 67.53.7. Let $S$ be a scheme. Let $f : Y \to X$ be a universally injective, integral morphism of algebraic spaces over $S$.
The functor
is fully faithful and its essential image is those sheaves of sets $\mathcal{F}$ on $X_{\acute{e}tale}$ whose restriction to $|X| \setminus f(|Y|)$ is isomorphic to $*$, and
the functor
is fully faithful and its essential image is those abelian sheaves on $Y_{\acute{e}tale}$ whose support is contained in $f(|Y|)$.
\[ f_{small, *} : \mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale}) \longrightarrow \mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale}) \]
\[ f_{small, *} : \textit{Ab}(Y_{\acute{e}tale}) \longrightarrow \textit{Ab}(X_{\acute{e}tale}) \]
In both cases $f_{small}^{-1}$ is a left inverse to the functor $f_{small, *}$.
Proof. Since $f$ is integral it is universally closed (Lemma 67.45.7). In particular, $f(|Y|)$ is a closed subset of $|X|$ and the statements make sense. The rest of the proof is identical to the proof of Lemma 67.13.5 except that we use Étale Cohomology, Proposition 59.47.1 instead of Étale Cohomology, Proposition 59.46.4. $\square$
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Comment #1794 by Matthieu Romagny on