Let $n \ge 1$ an integer. We consider the polynomial map $f: \mathbb{C} \to \mathbb{C}, z \mapsto z^n$.
Questions:
I) is this map $f$ proper for every $n$ (in topological sense) and why?
II) Let $r >0$ arbitrary positive real number. Consider the open disc (=open 1D ball) $B_r(0)$ with radius $r$ abound $0$.
Is the restricted map $f: B_r(0) \to B_{r^n}(0), z \mapsto z^n$ proper?
Attention: we consider restriction of $f$ between open discs; for their closures the question is trivial since the closures are compact.
Recall: In topology $f$ is called proper if for every compact $K$ set the preimage $f^{-1}(K)$ is also compact.
EDIT #1/ideas: I know a criterion that says that a map is proper if it is closed & it's fibers are compact. What do we know about $f$? The fibers are obviously finite (= zeros of $z^n-a$) and thus compact. I don't know if $f$ is closed. Since we are working with complex numbers $f$ is definitely surjective in I) as well in II) by choice of radius. Does it help? Possibly there is another strategy to show properness.
EDIT#2: Assume we know I) is true (polynomial maps are closed & fibrs finite thus proper. Can we conclude for this that II) is also proper? The main obstruction I see is that the (open) inclusion of the disc $B_r(0) \subset \mathbb{C}$ is obviuosly not proper. Thus a composition argument fails.
Best Answer
As the comments got too long, I will make them in an answer. Let $P$ a nonconstant polynomial in the plane. Then:
1: $P$ is proper as a map $P:\mathbb{C} \to \mathbb{C}$
2: If we restrict $P$ to a domain $D$ and $\Omega=P(D)$ is the image domain ($P$ is an open map being analytic nonconstant so $\Omega$ is open, while it is connected by continuity), then $P$ is proper as a map $P: D \to \Omega$ if and only if $P(\partial D) \cap \Omega =\emptyset$
Proof: $|P(z)| \to \infty$ as $|z| \to \infty$ hence the preimage of bounded sets by $P$ is bounded. But in the plane, $K$ compact iff $K$ is bounded and closed and obviously $P^{-1}(K)$ is then bounded by the above argument and closed by continuity, hence it is compact, hence $P$ is proper as a plane map.
For 2, we notice that if $K \subset \Omega$, $K$ compact, then $P^{-1}(K) \cap D=L$ is bounded since $P^{-1}(K)$ is so, but $L$ is only relatively closed in $D$. However if $P(\partial D) \cap \Omega =\emptyset$ and $x_n \in L, x_n \to x$, $P(x_n) \to P(x) \in K$ so $x \in \bar D, P(x) \in \Omega$. As $\bar D = D \cup \partial D$ and we cannot have $x \in \partial D$ by hypothesis, it follows $x \in D$, hence $x \in L$, hence $L$ closed, so compact.
Conversely, if $P$ restricted to $D$ is proper and $x \in \partial D, P(x)=y \in \Omega$, then $P(z)=(z-x)^nQ(z)+y, n \ge 1, Q(x) \ne 0$ hence the preimage of a small closed disc centered at $y$ and included in $\Omega$ has a component that is a a compact neighborhood of $x$ in the plane, hence its intersection with $D$ is not compact which contradicts the fact that $P$ is proper restricted to $D$.
A simple example to show that this happens is $P(z)=z(z-1)$ on the open unit disc. Clearly $0 \in P(D)$ but a small closed disc centered at zero has as preimage in the open disc, two components, one around zero compact, but one that is the intersection of a compact neighborhood of $1$ with the open unit disc, hence non-compact.
In the examples of the OP, $z \to z^n$ on the discs centered at the origin, obviously the boundary condition is satisfied, so the maps are proper when restricted too, but if we shift the domain to a disc containing $1$ but having another root of unit of order $n$ on the boundary, the restricted maps are not proper any more