**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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