Prove that there is no proper holomorphic map \[ f:\Delta\to\mathbb C. \]
Solution
Suppose, for contradiction, that \[ f:\Delta\to\mathbb C \] is proper and holomorphic.
1. The function \(f\) is nonconstant.
If \(f\equiv c\), then the compact set \(\{c\}\subset\mathbb C\) has preimage \[ f^{-1}(\{c\})=\Delta, \] which is not compact. Thus a constant map cannot be proper.
2. Properness implies\[ |f(z)|\longrightarrow\infty \qquad\text{uniformly as }|z|\to1. \]
Fix \(R>0\). The closed disk \[ K_R=\{w\in\mathbb C:|w|\le R\} \] is compact. Since \(f\) is proper, \(f^{-1}(K_R)\) is compact in \(\Delta\). Therefore there is some \(r_R<1\) such that \[ f^{-1}(K_R)\subset\{|z|\le r_R\}. \] Hence \[ |z|>r_R\quad\Longrightarrow\quad |f(z)|>R. \] This is precisely the asserted uniform divergence at the boundary.
3. The zero set of \(f\) in \(\Delta\) is finite.
The fiber \[ f^{-1}(0) \] is compact by properness. Since \(f\) is nonconstant by 1, its zeros are isolated. A compact discrete subset of the disk is finite. Write the zeros, with multiplicities, as \[ a_1,\ldots,a_N. \] If there are no zeros, take \(N=0\).
4. Define\[ P(z)=\prod_{j=1}^N(z-a_j) \]and\[ g(z)=\frac{P(z)}{f(z)}. \]Then \(g\) extends to a holomorphic, nonzero function on all of \(\Delta\).
The list \(a_1,\ldots,a_N\) repeats each zero according to its multiplicity, so \(P\) vanishes to exactly the same order as \(f\) at every zero of \(f\). Thus each apparent singularity of \(P/f\) is removable, and \(g\) extends holomorphically across all zeros of \(f\).
After cancellation, the extended value at each zero is nonzero; away from the zeros both numerator and denominator are nonzero. Hence \(g\) is nowhere zero on \(\Delta\).
5. One has\[ g(z)\longrightarrow0 \qquad\text{uniformly as }|z|\to1. \]
For \(z\in\Delta\), \[ |z-a_j|\le |z|+|a_j|<2, \] so \[ |P(z)|\le2^N. \] By 2, \(|f(z)|\to\infty\) uniformly as \(|z|\to1\). Therefore \[ |g(z)|=\frac{|P(z)|}{|f(z)|} \le\frac{2^N}{|f(z)|} \longrightarrow0 \] uniformly near the boundary.
6. The maximum modulus principle forces \(g\equiv0\), a contradiction.
Fix \(z_0\in\Delta\) and let \(\varepsilon>0\). By 5 there is \(r_0<1\) such that \[ |g(z)|<\varepsilon \qquad(r_0<|z|<1). \] Choose \(r\) with \[ \max\{r_0,|z_0|\}<r<1. \] On the disk \(|z|\le r\), the maximum modulus principle gives \[ \max_{|z|\le r}|g(z)| =\max_{|z|=r}|g(z)| <\varepsilon. \] In particular \(|g(z_0)|<\varepsilon\). Since \(\varepsilon\) is arbitrary, \(g(z_0)=0\). As \(z_0\) was arbitrary, \(g\equiv0\) on \(\Delta\), contradicting 4.
Therefore no proper holomorphic map \(\Delta\to\mathbb C\) exists.