The Stacks project

Proof. This is immediate from the definition. $\square$

Proof. Omitted. $\square$

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. $\square$

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. $\square$

Proof. Omitted. $\square$

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. $\square$

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. $\square$