Complex polynomials: preimage of a large enough disc contains the disc compactly?

I'm studying a proof (by Hubbard) of an elementary result in complex dynamics: if a complex polynomial $p$ of degree at least 2 has all of its critical points in its filled Julia set $K_p$, then $K_p$ is connected.

Hubbard's proof relies on the following elementary assertion: there is a sufficiently large $R > 0$ such that the disc $D_R(0)$ is contained inside $p^{-1}(D_R(0))$ as a compact subset. He uses this to construct an increasing sequence of sets whose union covers $K_p^c$, and then uses some basic topological facts about covering maps to prove that $K_p^c$ is homeomorphic to the annulus, implying $K_p$ is connected.

Anyway, I'm having trouble proving the assertion previously stated. I find it more reasonable to believe the opposite inclusion, that is $D_R \subset p(D_R)$ for large $R$. I say this on the basis that applying the polynomial map of degree $k$ will affect a boundary point of a large disc on the order of $a_k |z|^k$, in particular stretching it to a point far outside the disc.

But perhaps there's something elementary that I'm missing, like an easy application of a particular classical theorem in complex analysis?

• Clearly, this is false as stated: for the polynomial $z\mapsto z^2$ the preimage of $D_R(0)$ does not contain $D_R(0)$ when $R>1$. Can you provide more context or reference to the original? – you can call me Al Jan 28 '14 at 4:37
• Great, thanks for the reassurance. I talked to him about it and it did turn out to be a typo, which will be amended. I'm just glad I'm not as unprepared as I thought... – ithmath Jan 28 '14 at 17:04