Existence of $f\colon U\to V$ holomorphic non constant

Question

Let $Y$ be a complex mainfold and $\Omega\subseteq\Bbb C^n$ domain, $n<\dim Y$. Take $z_0\in\Omega$ and $y_0\in Y$.

I need a non-constant holomorphic mapping $f\colon U\to V$ where $U\subset Y$ and $V\subset\Omega$ are open neighborhood of $y_0$ and $z_0$ respectively, such that $f(y_0)=z_0$ and $f(y)\neq z_0$ for all $y\neq z_0$. Does such a mapping exist? It seems straightforward that such a mapping does exist, but maybe there is something I don't see.

Answer 1

As expected, the answer is negative when $n \neq \dim{Y}$.

We can assume that $Y$ is an open subset of $\mathbb{C}^m$ with $m > n$ and $y=z_0=0$, and $f=(f_1,\ldots,f_n)$. By (II.4.22 – page 97) in https://www-fourier.ujf-grenoble.fr/%7Edemailly/manuscripts/agbook.pdf it follows that the maximal ideal $\mathfrak{m}$ of the ring $\mathcal{O}$ of stalks of holomorphic functions on $Y$ at $y$ is the nilradical of the ideal generated by $(f_1,\ldots,f_n)$.

Now, by (II.2.7 – page 81) in the same reference, $\mathcal{O}$ is local Noetherian. By Stacks, Prop 10.60.9 (and some algebraic nonsense) it follows that $\mathcal{O}/\mathfrak{m}^k$ has complex dimension $O(n^k)$.

But it is easy to see (using its description as a ring of power series with nonzero radius) that the complex dimension of $\mathcal{O}/\mathfrak{m}^k$ is equivalent to a multiple of $m^k$. Hence a contradiction.

Answer 2

Here is a proof different from the one written by Mindlack.

1. First of all, the case when $n=1$ is much easier to handle, the nonexistence of such $f$ is immediate from the Weierstrass Preparation Theorem for holomorphic functions. The same theorem implies that the local codimension of $f^{-1}(0)$ is at most 1 everywhere in $f^{-1}(0)$.

2. The case when $n\ge 2$ is more subtle. For instance, it follows from the following theorem about dimensions of analytic subsets:

Theorem. Suppose that $F=(f_1,...,f_n): M\to {\mathbb C}^n$ is holomorphic function on a complex manifold $M$. Then for every point $a\in F^{-1}(0)$, $$ codim_a F^{-1}(0) \le \sum_{j=1}^n codim_a f_j^{-1}(0). $$ Here $codm_a$ is the local codimension of an analytical subset of $M$.

See for instance Chapter 1, Section 3.5, Proposition 3 in

Chirka, E. M., Complex analytic sets. Translated from the Russian by R. A. M. Hoksbergen, Mathematics and Its Applications: Soviet Series, 46. Dordrecht etc.: Kluwer Academic Publishers. xix, 372 p. (1989). ZBL0683.32002.

Since $codim_a f_j^{-1}(0)$ is at most $1$ (see item 1 above), it follows that $$ dim_a F^{-1}(0) \ge dim(M)-n$$ (see corollary in Chirka's book that immediately follows the cited proposition). Hence, if $dim(M)>n$, $F^{-1}(0)$ cannot be a discrete subset of $M$.