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$ ?
 A: 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:


*

*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.

*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.
