Preimage of circle with critical point and its image inside is simple and closed

Question

I am reading Devaney's "An Introduction to Chaotic Dynamical Systems" and I am trying to convince myself of a claim made there (Section 3.6, proposition 6.2). The proposition concerns showing that the Julia set of $f(z)=z^2+c$ is simple and closed if $|c|<1/4$. The claim is

If the critical point $0$ and its image $c$ is contained in a circle $\Gamma$ then the preimage of $\Gamma$ is a simple closed curve.

I am stuck on how to prove this. Would I need to construct a continuous limiting curve as in the proof for the Julia set?

Answer 1

In fact, we can write down a pretty simple parameterization for this simple closed curve. Suppose that our circle $\Gamma$ is parametrized by $$ \gamma(t) = z_0 + r e^{it}, \text{ for } \: 0\leq t \leq 2\pi. $$ Then, the parameterization of $f^{-1}(\Gamma)$ will be $$ p(t) = \pm \sqrt{(z_0-c) + r e^{it}}. $$ The fact that $c$ is inside $\Gamma$ implies that the shifted circle $\Gamma_c = (z_0-c) + r e^{it}$ contains the origin. Thus it boils down to showing that the inverse image of any circle under the square function contains the origin. This is a simple consequence of the fact that the square root function maps the circle to the right half-plane, the negative square root function maps the circle to the left half-plane, and the endpoints match up.