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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