Proper holomorphic map

Remarks

Equivalent form: $F$ is proper if and only if it has the following property. For every sequence $(z_k)$ in $\Omega_1$ that eventually leaves every compact subset of $\Omega_1$, the sequence $(F(z_k))$ eventually leaves every compact subset of $\Omega_2$. For bounded domains this says that $\operatorname{dist}(F(z), \partial\Omega_2) \to 0$ as $\operatorname{dist}(z, \partial\Omega_1) \to 0$.

Example: let $\mathbb{D}$ be the unit disc and $F(z) = z^k$ with $k \ge 1$. A compact $K \subset \mathbb{D}$ lies in a closed disc $\set{\abs{w} \le r}$ with $r < 1$, and $F^{-1}(K)$ is a closed subset of $\set{\abs{z} \le r^{1/k}}$, hence compact. The map $(z_1, z_2) \mapsto (z_1, z_2^2)$ of the unit polydisc in $\C^2$ to itself is proper for the same reason.

Non-example: the inclusion $F \colon \set{\abs{z} < 1/2} \to \mathbb{D}$ is holomorphic but not proper. For the compact set $K = \set{\abs{w} \le 1/2}$, the preimage $F^{-1}(K) = \set{\abs{z} < 1/2}$ is not compact.

A proper holomorphic map between domains of the same dimension $n$ is surjective, its fibres are finite, and it is a finite branched covering. A biholomorphism is a proper holomorphic map. The proper holomorphic self-maps of $\mathbb{D}$ are exactly the finite Blaschke products.