# Show that there is no proper holomorphic map from an annulus $A_r=\{z \in \mathbb C:1 <|z| < r \}$ to the punctured unit disc

Show that there is no proper holomorphic map from an annulus $A_r=\{z \in \mathbb C:1 <|z| < r \}$ to the punctured unit disc.

Def:A map $f: X \to Y$ is called proper if $f^{-1}(K)$ is compact for every compact set $K$ in Y.

Author gives a hint that there is a covering map from unit disc to punctured unit disc whose fibers are infinite.But I am unable to see how to use this hint. please give some hints/ideas to prove this.

The non existence of such a map in the reverse direction can be found here:Non Existence of a proper holomorphic map from the punctured unit disc to an Annulus.

Can someone please give a reference for reading about construction of proper maps between different domains in $\mathbb C$ ?

## 1 Answer

Here's one way to prove this, although it doesn't use the hint.

Let $\mathbb{D}^*$ denote the punctured unit disk, and suppose that $f\colon A_r \to \mathbb{D}^*$ is a proper holomorphic map. I claim that the limits $$\lim_{|z|\to 1^+} |f(z)| \qquad\text{and}\qquad \lim_{|z|\to r^-} |f(z)|$$ both exist, and are each either $0$ or $1$.

To see this, let $\epsilon > 0$, and consider the map $g\colon A_r \to (0,1)$ defined by $g(z) = |f(z)|$. Note that $g$ is proper, being the composition of two proper maps. Then $K = g^{-1}\bigl([\epsilon,1-\epsilon]\bigr)$ is a compact subset of $A_r$. By the tube lemma, there exists a $\delta >0$ so that $A_{1+\delta} \subset A_r -K$, where $A_{1+\delta}$ denotes the annulus $1<|z|<1+\delta$, so $g$ maps $A_{1+\delta}$ to the complement of $[\epsilon,1-\epsilon]$. Since $A_{1+\delta}$ is connected, it follows that either $g(A_{1+\delta}) \subset (0,\epsilon)$ or $g(A_{1+\delta})\subset (1-\epsilon,1)$. Since $\epsilon$ was arbitrary, we conclude that the first limit exists, and is equal to either $0$ or $1$. A similar argument holds for the second limit.

We now obtain a contradiction, using two cases:

1. Suppose one of the limits above is equal to $0$, say the first limit. In this case we get a contradiction from the Schwarz reflection principle. Specifically, let $R$ be the region $0<\mathrm{Im}(z)<\log(r)$, and let $h\colon R\to\mathbb{C}$ be the function $h(z) = f\bigl(e^{-iz}\bigr)$. Then $R$ is holomorphic and tends to zero as $\mathrm{Im}(z) \to 0$. By the Schwarz reflection principle, $h$ extends to a holomorphic function on the region $-\log(r) < \mathrm{Im}(z) < \log(r)$. But $h$ is $0$ on the real axis, so $h$ must be the zero function, a contradiction.

2. Suppose that both of the limits above are equal to $1$. In this case we get a contradiction from the minimum modulus principle. Let $\epsilon > 0$. Since $|f(z)| \to 1$ as $z \to \partial A_r$, there exists a $\delta > 0$ so that $|f(z)| > 1-\epsilon$ when $z$ is within $\delta$ of $\partial A_r$. Using the minimum modulus principle for $f$ on the annulus $1+\delta \leq |z| \leq r-\delta$, we conclude that $|f(z)| > 1-\epsilon$ for all $z\in A_r$, a contradiction since $\epsilon$ was arbitrary.

• Sorry..can you please explain how are you using tube lemma? – Dontknowanything Jun 4 '15 at 9:56
• @Learner If $U$ is an open set in $[0,1]\times S^1$ containing $\{0\}\times S^1$, then there exists a "tube" $[0,\epsilon) \times S^1$ contained in $U$. – Jim Belk Jun 4 '15 at 14:39