Definition 29.45.1. A morphism f : X \to Y of schemes is called a universal homeomorphism if the base change f' : Y' \times _ Y X \to Y' is a homeomorphism for every morphism Y' \to Y.
29.45 Universal homeomorphisms
The following definition is really superfluous since a universal homeomorphism is really just an integral, universally injective and surjective morphism, see Lemma 29.45.5.
First we state the obligatory lemmas.
Lemma 29.45.2. The base change of a universal homeomorphism of schemes by any morphism of schemes is a universal homeomorphism.
Proof. This is immediate from the definition. \square
Lemma 29.45.3. The composition of a pair of universal homeomorphisms of schemes is a universal homeomorphism.
Proof. Omitted. \square
The following simple lemma is the key to characterizing universal homeomorphisms.
Lemma 29.45.4. Let f : X \to Y be a morphism of schemes. If f is a homeomorphism onto a closed subset of Y then f is affine.
Proof. Let y \in Y be a point. If y \not\in f(X), then there exists an affine neighbourhood of y which is disjoint from f(X). If y \in f(X), let x \in X be the unique point of X mapping to y. Let y \in V be an affine open neighbourhood. Let U \subset X be an affine open neighbourhood of x which maps into V. Since f(U) \subset V \cap f(X) is open in the induced topology by our assumption on f we may choose a h \in \Gamma (V, \mathcal{O}_ Y) such that y \in D(h) and D(h) \cap f(X) \subset f(U). Denote h' \in \Gamma (U, \mathcal{O}_ X) the restriction of f^\sharp (h) to U. Then we see that D(h') \subset U is equal to f^{-1}(D(h)). In other words, every point of Y has an open neighbourhood whose inverse image is affine. Thus f is affine, see Lemma 29.11.3. \square
Lemma 29.45.5. Let f : X \to Y be a morphism of schemes. The following are equivalent:
f is a universal homeomorphism, and
f is integral, universally injective and surjective.
Proof. Assume f is a universal homeomorphism. By Lemma 29.45.4 we see that f is affine. Since f is clearly universally closed we see that f is integral by Lemma 29.44.7. It is also clear that f is universally injective and surjective.
Assume f is integral, universally injective and surjective. By Lemma 29.44.7 f is universally closed. Since it is also universally bijective (see Lemma 29.9.4) we see that it is a universal homeomorphism. \square
Lemma 29.45.6. Let X be a scheme. The canonical closed immersion X_{red} \to X (see Schemes, Definition 26.12.5) is a universal homeomorphism.
Proof. Omitted. \square
Lemma 29.45.7. Let f : X \to S and S' \to S be morphisms of schemes. Assume
S' \to S is a closed immersion,
S' \to S is bijective on points,
X \times _ S S' \to S' is a closed immersion, and
X \to S is of finite type or S' \to S is of finite presentation.
Then f : X \to S is a closed immersion.
Proof. Assumptions (1) and (2) imply that S' \to S is a universal homeomorphism (for example because S_{red} = S'_{red} and using Lemma 29.45.6). Hence (3) implies that X \to S is homeomorphism onto a closed subset of S. Then X \to S is affine by Lemma 29.45.4. Let U \subset S be an affine open, say U = \mathop{\mathrm{Spec}}(A). Then S' = \mathop{\mathrm{Spec}}(A/I) by (1) for a locally nilpotent ideal I by (2). As f is affine we see that f^{-1}(U) = \mathop{\mathrm{Spec}}(B). Assumption (4) tells us B is a finite type A-algebra (Lemma 29.15.2) or that I is finitely generated (Lemma 29.21.7). Assumption (3) is that A/I \to B/IB is surjective. From Algebra, Lemma 10.126.9 if A \to B is of finite type or Algebra, Lemma 10.20.1 if I is finitely generated and hence nilpotent we deduce that A \to B is surjective. This means that f is a closed immersion, see Lemma 29.2.1. \square
Lemma 29.45.8. Let f : X \to Z be the composition of two morphisms g : X \to Y and h : Y \to Z. If two of the morphisms \{ f, g, h\} are universal homeomorphisms, so is the third morphism.
Proof. If both of g and h are universal homeomorphisms, so is f by Lemma 29.45.3.
Suppose both of f and g are universal homeomorphisms. We want to show that h is also. Now base change the diagram along an arbitrary morphism \alpha : Z' \to Z of schemes, we get the following diagram with all squares Cartesian:
Our assumption implies that the composition f'= h' \circ g' : X' \to Z' and g' : X' \to Y' are homeomorphisms, therefore so is h'. This finishes the proof of h being a universal homeomorphism.
Finally, assume f and h are universal homeomorphisms. We want to show that g is a universal homeomorphism. Let \beta : Y' \to Y be an arbitrary morphism of schemes. We get the following diagram with all squares Cartesian:
Here the morphism \gamma : Y' \to Y'' is defined by the universal property of fiber products and the two morphisms id_{Y'} : Y' \to Y' and \beta : Y' \to Y. We shall prove that g' is a homeomorphism. Since the property of being a homeomorphism has 2-out-of-3 property, we see that g'' is a homeomorphism. Staring at the top square, it suffices to prove that \gamma is a universal homeomorphism. Since h'' is a homeomorphism, we see that it is an affine morphism by Lemma 29.45.4 and a fortiori separated (Lemma 29.11.2). Since h'' \circ \gamma is the identity, we see that \gamma is a closed immersion by Schemes, Lemma 26.21.11. Since h'' is bijective, it follows that \gamma is a bijective closed immersion and hence a universal homeomorphism (for example by the characterization in Lemma 29.45.5) as desired. \square
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