# Polynomials mapping surjectively onto components

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 $$z_0\in D'$$.

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 $$q(D') = D$$.