Lemma 65.49.8. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$ satisfying the equivalent conditions of Lemma 65.49.1.

1. The normalization $X^\nu$ is normal.

2. The morphism $\nu : X^\nu \to X$ is integral and surjective.

3. The map $|\nu | : |X^\nu | \to |X|$ induces a bijection between the sets of points of codimension $0$ (Properties of Spaces, Definition 64.10.2).

4. Let $Z \to X$ be a morphism. Assume $Z$ is a normal algebraic space and that for $z \in |Z|$ we have: $z$ has codimension $0$ in $Z \Rightarrow f(z)$ has codimension $0$ in $X$. Then there exists a unique factorization $Z \to X^\nu \to X$.

Proof. Properties (1), (2), and (3) follow from the corresponding results for schemes (Morphisms, Lemma 29.54.5) combined with the fact that a point of a scheme is a generic point of an irreducible component if and only if the dimension of the local ring is zero (Properties, Lemma 28.10.4).

Let $Z \to X$ be a morphism as in (4). Let $U$ be a scheme and let $U \to X$ be a surjective étale morphism. Choose a scheme $V$ and a surjective étale morphism $V \to U \times _ X Z$. The condition on codimension $0$ points assures us that $V \to U$ maps generic points of irreducible components of $V$ to generic points of irreducible components of $U$. Thus we obtain a unique factorization $V \to U^\nu \to U$ by Morphisms, Lemma 29.54.5. The uniqueness guarantees us that the two maps

$V \times _{U \times _ X Z} V \to V \to U^\nu$

agree because these maps are the unique factorization of the map $V \times _{U \times _ X Z} V \to V \to U$. Since the algebraic space $U \times _ X Z$ is equal to the quotient $V/V \times _{U \times _ X Z} V$ (see Spaces, Section 63.9) we find a canonical morphism $U \times _ X Z \to U^\nu$. Picture

$\xymatrix{ U \times _ X Z \ar[r] \ar[d] & U^\nu \ar[r] \ar[d] & U \ar[d] \\ Z \ar@/_/[rr] \ar@{..>}[r] & X^\nu \ar[r] & X }$

To obtain the dotted arrow we note that the construction of the arrow $U \times _ X Z \to U^\nu$ is functorial in the étale morphism $U \to X$ (precise formulation and proof omitted). Hence if we set $R = U \times _ X U$ with projections $s, t : R \to U$, then we obtain a morphism $R \times _ X Z \to R^\nu$ commuting with $s, t : R \to U$ and $s^\nu , t^\nu : R^\nu \to U^\nu$. Recall that $X^\nu = U^\nu /R^\nu$, see proof of Lemma 65.49.5. Since $X = U/R$ a simple sheaf theoretic argument shows that $Z = (U \times _ X Z)/(R \times _ X Z)$. Thus the morphisms $U \times _ X Z \to U^\nu$ and $R \times _ X Z \to R^\nu$ define a morphism $Z \to X^\nu$ as desired. $\square$

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