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 65.9) 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 67.49.5. 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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