Non Existence of a proper holomorphic map from the unit disc onto the complex plane It is well known that there is no proper holomorphic map from complex plane onto disc by Liouville's theorem.Does there exist a proper holomorphic map $f$ from the unit disc onto the complex plane?I believe that such map does not exists but I'm unable to prove this.Please Help! 
Def:A map $f:X \to Y$ is called a proper map if $f^{-1}(K)$ is compact in $X$ for every compact set $K$ in $Y$.
 A: There is no such map.  For convenience, I will use the upper half-plane $\mathbb{H}$ instead of the unit disk.
Suppose $f\colon \mathbb{H}\to\mathbb{C}$ is a proper holomorphic map.
Note first that, for each sequence $\{z_n\}$ in $\mathbb{H}$ that converges to a point $p$ on the real axis, the image sequence $\{f(z_n)\}$ must must converge to $\infty$ on the Riemann sphere.  To prove this, observe that if $\overline{D}_R$ is the closed disk of radius $R$ in $\mathbb{C}$ centered at the origin, then $\overline{D}_R$ is compact, and hence $f^{-1}(\overline{D}_R)$ is a compact subset of $\mathbb{H}$.  Then $\mathbb{C}-f^{-1}(\overline{D}_R)$ is an open neighborhood of $p$, so $z_n$ lies in the complement of $f^{-1}(\overline{D}_R)$ for all but finitely many $n$.  This holds for all $R>0$, which proves the claim.
Now, since $f$ is a nonzero holomorphic function, the zeroes of $f$ form a discrete subset of $\mathbb{H}$.  Since $f$ is proper, the set of zeroes must be compact, and therefore there are only finitely many zeroes.  Let $D$ be any open disk centered on the real axis that does not contain any zeroes of $f$, and consider the holomorphic function $g(z) = 1/f(z)$ on $D\cap\mathbb{H}$.
Now, $g$ has the property that $g(z_n) \to 0$ for any sequence $z_n \in D\cap\mathbb{H}$ converging to a point on the real axis. Thus $g$ extends continuously to $D\cap\overline{\mathbb{H}}$ by setting $g(x) = 0$ for $x \in D\cap\mathbb{R}$.  By the Schwarz reflection principle, $g$ now extends to a holomorphic function on all of $D$, which is nonsense since $g$ is zero on $D\cap\mathbb{R}$.
