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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)