Lemma 29.47.7. Let $X$ be a scheme.

1. The category of universal homeomorphisms $Y \to X$ has an initial object $X^{awn} \to X$.

2. Given $Y \to X$ in the category of (1) the resulting morphism $X^{awn} \to Y$ is an isomorphism if and only if $Y$ is absolutely weakly normal.

3. The category of universal homeomorphisms $Y \to X$ which induce ismomorphisms on residue fields has an initial object $X^{sn} \to X$.

4. Given $Y \to X$ in the category of (3) the resulting morphism $X^{sn} \to Y$ is an isomorphism if and only if $Y$ is seminormal.

For any morphism $h : X' \to X$ of schemes there are unique morphisms $h^{awn} : (X')^{awn} \to X^{awn}$ and $h^{sn} : (X')^{sn} \to X^{sn}$ compatible with $h$.

Proof. We will prove (3) and (4) and omit the proof of (1) and (2). Let $h : X' \to X$ be a morphism of schemes. If (3) holds for $X$ and $X'$, then $X' \times _ X X^{sn} \to X'$ is a universal homeomorphism and hence we get a unique morphism $(X')^{sn} \to X' \times _ X X^{sn}$ over $X'$ by the universal property of $(X')^{sn} \to X'$. Composed with the projection $X' \times _ X X^{sn} \to X^{sn}$ we obtain $h^{sn}$. If in addition (4) holds for $X$ and $X'$ and $h$ is an open immersion, then $X' \times _ X X^{sn}$ is seminormal (Lemma 29.47.4) and we deduce that $(X')^{sn} \to X' \times _ X X^{sn}$ is an isomorphism.

Recall that any universal homeomorphism is affine, see Lemma 29.45.4. Thus if $X$ is affine then (3) and (4) follow immediately from Lemma 29.47.6. Let $X$ be a scheme and let $\mathcal{B}$ be the set of affine opens of $X$. For each $U \in \mathcal{B}$ we obtain $U^{sn} \to U$ and for $V \subset U$, $V, U \in \mathcal{B}$ we obtain a canonical isomorphism $\rho _{V, U} : V^{sn} \to V \times _ U U^{sn}$ by the discussion in the previous paragraph. Thus by relative glueing (Constructions, Lemma 27.2.1) we obtain a morphism $X^{sn} \to X$ which restricts to $U^{sn}$ over $U$ compatibly with the $\rho _{V, U}$. Next, let $Y \to X$ be a universal homeomorphism. Then $U \times _ X Y \to U$ is a universal homeomorphism for $U \in \mathcal{B}$ and we obtain a unique morphism $g_ U : U^{sn} \to U \times _ X Y$ over $U$. These $g_ U$ are compatible with the morphisms $\rho _{V, U}$; details omitted. Hence there is a unique morphism $g : X^{sn} \to Y$ over $X$ agreeing with $g_ U$ over $U$, see Constructions, Remark 27.2.3. This proves (3) for $X$. The proof of (4) is similar; details omitted. $\square$

## Comments (2)

Comment #4306 by correction_bot on

The proof says we'll only prove (1) and (2) but then talks about $X^{sn}$ which only plays a role in (3) and (4) (should be $X^{awn}$ instead).

## 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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EUS. Beware of the difference between the letter 'O' and the digit '0'.