TITLE: What is the universal property of normalization?
QUESTION [31 upvotes]: What is the universal property of normalization? I'm looking for an answer something like

If X is a scheme and Y→X is its
  normalization, then the morphism
  Y→X has property P and any other
  morphism Z→X with property P
  factors uniquely through Y.

REPLY [3 votes]: While this comes pretty late I wanted to add that normalization has a somewhat stronger universal property than what is mentioned in the other answers:
(N) Let $f: Z \to X$ be a morphism between integral, excellent schemes with $Z$ normal such that the image contains a normal point of $X$. Then $f$ lifts uniquely to the normalization $Y \to X$.
If the cited normal point is the generic point of $X$, one gets back the criterion demanding dominance (ND), as in Giulia's answer. However it also encompasses situations like a normal curve $Z$ on a singular surface $X$ not contained in the singular locus, which also lifts uniquely to the normalization. 
To justify (N) one uses (ND) to reduce to $Z$ being the normalization of a closed integral subscheme $R \subset X$ whose generic point $\eta$ is normal in $X$. Denote by $R'$ the reduced closure of $\{\eta\}$ in the normalization of $R \times_X Y$. Because $\eta$ is a normal point of $X$, this is the unique component mapping dominantly to $R$. In the resulting diagram
$\require{AMScd}$
$$\begin{CD}
R' @>>> R\times_X Y@>>> Y\\
@VVV @VVV @VVV\\
Z @>>> R@>>> X
\end{CD}$$
we get the left vertical arrow again by (ND) from $R' \to R$. However, as $X$ is excellent, the map $Y \to X$ is finite, hence $R' \to Z$ is a finite birational map to a normal scheme, so an isomorphism. This gives a lift $Z \to R \times_X Y$ in the above diagram. Any other such lift also induces a section of $R' \to Z$ by (ND) which implies uniqueness.