The Stacks project

Lemma 28.45.7. 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 (1) and (2) and omit the proof of (3) and (4). Let $h : X' \to X$ be a morphism of schemes. If (1) 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 (2) holds for $X$ and $X'$ and $h$ is an open immersion, then $X' \times _ X X^{sn}$ is seminormal (Lemma 28.45.4) and we deduce that $(X')^{sn} \to X' \times _ X X^{sn}$ is an isomorphism.

Recall that any universal homeomorphism is affine, see Lemma 28.43.4. Thus if $X$ is affine then (1) and (2) follow immediately from Lemma 28.45.6. 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 26.2.1) 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 26.2.3. This proves (1) for $X$. The proof of (2) is similar; details omitted. $\square$

Comments (1)

Comment #4306 by correction_bot on

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

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