Lemma 66.12.1. Let $S$ be a scheme. Let $Z \to X$ be an immersion of algebraic spaces. Then $|Z| \to |X|$ is a homeomorphism of $|Z|$ onto a locally closed subset of $|X|$.

## 66.12 Reduced spaces

We have already defined reduced algebraic spaces in Section 66.7. Here we just prove some simple lemmas regarding reduced algebraic spaces.

**Proof.**
Let $U$ be a scheme and $U \to X$ a surjective étale morphism. Then $Z \times _ X U \to U$ is an immersion of schemes, hence gives a homeomorphism of $|Z \times _ X U|$ with a locally closed subset $T'$ of $|U|$. By Lemma 66.4.3 the subset $T'$ is the inverse image of the image $T$ of $|Z| \to |X|$. The map $|Z| \to |X|$ is injective because the transformation of functors $Z \to X$ is injective, see Spaces, Section 65.12. By Topology, Lemma 5.6.4 we see that $T$ is locally closed in $|X|$. Moreover, the continuous map $|Z| \to T$ is a homeomorphism as the map $|Z \times _ X U| \to T'$ is a homeomorphism and $|Z \times _ Y U| \to |Z|$ is submersive.
$\square$

The following lemma will help us construct (locally) closed subspaces.

Lemma 66.12.2. Let $S$ be a scheme. Let $j : R \to U \times _ S U$ be an étale equivalence relation. Let $X = U/R$ be the associated algebraic space (Spaces, Theorem 65.10.5). There is a canonical bijection

Moreover, if $Z \to X$ is closed (resp. open) if and only if $Z' \to U$ is closed (resp. open).

**Proof.**
Denote $\varphi : U \to X$ the canonical map. The bijection sends $Z \to X$ to $Z' = Z \times _ X U \to U$. It is immediate from the definition that $Z' \to U$ is an immersion, resp. closed immersion, resp. open immersion if $Z \to X$ is so. It is also clear that $Z'$ is $R$-invariant (see Groupoids, Definition 39.19.1).

Conversely, assume that $Z' \to U$ is an immersion which is $R$-invariant. Let $R'$ be the restriction of $R$ to $Z'$, see Groupoids, Definition 39.18.2. Since $R' = R \times _{s, U} Z' = Z' \times _{U, t} R$ in this case we see that $R'$ is an étale equivalence relation on $Z'$. By Spaces, Theorem 65.10.5 we see $Z = Z'/R'$ is an algebraic space. By construction we have $U \times _ X Z = Z'$, so $U \times _ X Z \to Z$ is an immersion. Note that the property “immersion” is preserved under base change and fppf local on the base (see Spaces, Section 65.4). Moreover, immersions are separated and locally quasi-finite (see Schemes, Lemma 26.23.8 and Morphisms, Lemma 29.20.16). Hence by More on Morphisms, Lemma 37.57.1 immersions satisfy descent for fppf covering. This means all the hypotheses of Spaces, Lemma 65.11.5 are satisfied for $Z \to X$, $\mathcal{P}=$“immersion”, and the étale surjective morphism $U \to X$. We conclude that $Z \to X$ is representable and an immersion, which is the definition of a subspace (see Spaces, Definition 65.12.1).

It is clear that these constructions are inverse to each other and we win. $\square$

Lemma 66.12.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a closed subset. There exists a unique closed subspace $Z \subset X$ with the following properties: (a) we have $|Z| = T$, and (b) $Z$ is reduced.

**Proof.**
Let $U \to X$ be a surjective étale morphism, where $U$ is a scheme. Set $R = U \times _ X U$, so that $X = U/R$, see Spaces, Lemma 65.9.1. As usual we denote $s, t : R \to U$ the two projection morphisms. By Lemma 66.4.5 we see that $T$ corresponds to a closed subset $T' \subset |U|$ such that $s^{-1}(T') = t^{-1}(T')$. Let $Z' \subset U$ be the reduced induced scheme structure on $T'$. In this case the fibre products $Z' \times _{U, t} R$ and $Z' \times _{U, s} R$ are closed subschemes of $R$ (Schemes, Lemma 26.18.2) which are étale over $Z'$ (Morphisms, Lemma 29.36.4), and hence reduced (because being reduced is local in the étale topology, see Remark 66.7.3). Since they have the same underlying topological space (see above) we conclude that $Z' \times _{U, t} R = Z' \times _{U, s} R$. Thus we can apply Lemma 66.12.2 to obtain a closed subspace $Z \subset X$ whose pullback to $U$ is $Z'$. By construction $|Z| = T$ and $Z$ is reduced. This proves existence. We omit the proof of uniqueness.
$\square$

Lemma 66.12.4. Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$. Let $Z \subset X$ be a closed subspace. Assume $Y$ is reduced. A morphism $f : Y \to X$ factors through $Z$ if and only if $f(|Y|) \subset |Z|$.

**Proof.**
Assume $f(|Y|) \subset |Z|$. Choose a diagram

where $U$, $V$ are schemes, and the vertical arrows are surjective and étale. The scheme $V$ is reduced, see Lemma 66.7.1. Hence $h$ factors through $a^{-1}(Z)$ by Schemes, Lemma 26.12.7. So $a \circ h$ factors through $Z$. As $Z \subset X$ is a subsheaf, and $V \to Y$ is a surjection of sheaves on $(\mathit{Sch}/S)_{fppf}$ we conclude that $X \to Y$ factors through $Z$. $\square$

Definition 66.12.5. Let $S$ be a scheme, and let $X$ be an algebraic space over $S$. Let $Z \subset |X|$ be a closed subset. An *algebraic space structure on $Z$* is given by a closed subspace $Z'$ of $X$ with $|Z'|$ equal to $Z$. The *reduced induced algebraic space structure* on $Z$ is the one constructed in Lemma 66.12.3. The *reduction $X_{red}$ of $X$* is the reduced induced algebraic space structure on $|X|$.

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