Polynomials mapping surjectively onto components

Question

Let $q$ be a polynomial of degree $d\geq 1$ and let $K$ be a compact subset of the complex plane. Let $D$ and $D'$ be the unbounded components of $\mathbb{C}\setminus K$ and $\mathbb{C}\setminus q^{-1}(K)$ respectively. I want to show that $q(D') = D$ and $q(\partial D') = \partial D$.

Now its easy to see htat $q^{-1}(K)$ is compact as it is closed and bounded. Furthermore $q(D')$ is an open connected set, as $q$ is an open mapping, and since it is an unbounded subset of $\mathbb{C}\setminus K$ we must have that $q(D')\subset D$. However how do I prove equality here?

Given $w_0\in D$ the fundamental theorem of algebra implies that $\exists z_0\in \mathbb{C}$ such that $q(z_0) = w_0$. However how do I show that $z_0\in D'$.

Answer 1

I realize this can be answered as follows: Assuming that $q(D')\neq D$ we can then find a point $\zeta\in \partial q(D')\cap D$. Pick a sequence $(z_n)$ in $D'$ such that $q(z_n)\rightarrow \zeta$. Then since $(z_n)$ must be bounded we can find a convergent subsequence, convering to say $z$. By continuity $q(z) = \zeta$ and also $z\in \partial D'$. Since $\zeta$ is an interior point of $D$ we can find a neighbourhood of $z$ mapping into $D$. This contradicts the fact that $z\in \partial D'$ and therefore $q(D') = D$.

We can argue similarly for the boundary case.