When is normalization functorial?

Question

Let $X$ and $Y$ be two irreducible, affine $\newcommand{\C}{\mathbb C}\C$-varieties. Let $f:X\to Y$ be a morphism. Denote by $u:\tilde X\to X$ and $v:\tilde Y\to Y$ their normalizations. Now, if $f$ is dominant, I get an induced morphism $\tilde f: \tilde X\to \tilde Y$ such that $v\circ\tilde f= f\circ u$. This follows, for instance, from Corollary 17.4.4 in the book of Tauvel and Yu.

Now what if $f$ is not dominant? I would suppose that one does not usually get such an induced $\tilde f$, right? However, are there other conditions on $f$ that imply the existence of such a lift? What about open or closed immersions?

Looking at it naively stopped me right in my tracks: The relation between the function fields of $X$ and $Y$ isn't really obvious to me when the morphism isn't dominant. I have some ideas, but they might lead nowhere or they have been worked out before, so I am asking here first. Thanks a lot in advance.

Answer 1

Closed immersions obviously don't work (think about the node, if I include the singular point into the node, where does it go in the normalization).

Open immersions are fine for obvious reasons (if it works on schemes, it works for open immersions of schemes).

EDIT: I'm going to make what I wrote here more precise.

In general I claim:

Theorem. $X \to Y$ induces $\tilde{X} \to \tilde{Y}$ if the generic point of $X$ is mapped to a point a of $Y$ where the normalization map $\tilde{Y} \to Y$ is an isomorphism.

Proof: Let $\eta$ be the generic point of $X$ with image $\gamma$ in $Y$. Set $Z$ to be the closure of $\gamma$ in $Y$ and $Z'$ the closure of $\gamma$ in $\tilde Y$ (note that the normalization of $Y$ is an isomorphism at $\gamma$ so this makes sense). Then the induced map $Z' \to Z$ is birational and is a finite map (since $\tilde Y \to Y$ is finite). It follows that we have a unique factorization $\tilde Z \to Z' \to Z$. Then as you already observed we have functoriality since $X \to Z$ is dominant.

That completes the proof.

Remark: I think this is pretty close to an if and only if, with one exception that I can think of. If $\gamma_{1}, \ldots, \gamma_{n}$ are the points of $\tilde{Y}$ mapping to $\gamma$, then we have $k(\gamma) \hookrightarrow k(\gamma_1) \oplus \ldots \oplus k(\gamma_n).$ We need there to be a unique set of ring maps $$ k(\gamma) \hookrightarrow k(\gamma_1) \oplus \ldots \oplus k(\gamma_n) \to k(\eta). $$ The only way that is possible I think is that if $n = 1$, and that there exist a unique embedding $k(\gamma_1) \subseteq k(\eta)$. I think that can only happen if $k(\gamma) \subseteq k(\gamma_1)$ is purely inseparable, and that inseparable extension happens to live inside $k(\eta)$.