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:

$\xymatrix{ X' \ar[r]^{g'} \ar[d] & Y' \ar[r]^{h'} \ar[d] & Z' \ar[d] \\ X \ar[r]^{g} & Y \ar[r]^{h} & Z. }$

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:

$\xymatrix{ X' \ar[r]^{g'} \ar[d] & Y' \ar[d]^{\gamma } & \\ X'' \ar[r]^{g''} \ar[d] & Y'' \ar[r]^{h''} \ar[d] & Y' \ar[d]^{h \circ \beta } \\ X \ar[r]^{g} & Y \ar[r]^{h} & Z. }$

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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