## open mapping theorem(complex analysis)

By Zorn leung

Does open mapping theorem holds in ${\mathbb{C}}^n$ for n>1?

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### 2 Responses to open mapping theorem(complex analysis)

1. KKK says:

Zorn:

What is the target space? If $f: \mathbb{C}^2 \rightarrow \mathbb{C}^2$. Trivially no. Example: $f(z, w)=(z, 0)$.

For $f: \mathbb{C}^2 \rightarrow \mathbb{C}$ (say). Trivially yes.
The reason is, let $U\subset \mathbb{C}^2$ be open and $p\in f(U)$. Let’s say $f(0, 0)=p$. Then it is easy to see that $f(z, 0)$ is nonconstant OR $f(0, w)$ is non-constant (by Taylor’s expansion, say). Say $g(z)=f(z, 0)$ is non-constant, then it maps a small (1-dim!) neighborhood of $0$ to an open set around $p$, by 1-D open mapping theorem. So there is an open set $V$ around $p$ which is contained in $f(U)$.
[My argument is wrong. ]

2. KKK says:

It turns out that I can correct my argument. Let my clarify it here.
For convenience, let’s assume $f: \mathbb{C}^2\rightarrow \mathbb{C}$ be a non-constant holomorphic (entire) function. Let $U\subset \mathbb{C}^2$ be open and $p\in f(U)$. Wlog, we assume $(0,0)\in U$ and $f(0, 0)=p$
. We claim that there is an open set $V\subset f(U)$ which contains $p$. To see this, since $f$ is non-constant, there is $(0, 0)\neq (a, b)\in \mathbb{C}^2$ with $f(a, b)\neq f(0, 0)$. Define $g(z):=f(az, bz)$ to be an entire non-constant holomorphic function. Then by open mapping theorem, $g$ maps arbitrary small neighborhood of $0$ onto an open set $V$ around $p$. So by choosing this neighborhood to be small enough, we see that there is an open set $V\subset f(U)$ containing $p$.