Let $f$ be an holomorphic function and $K(f)$ its non-escaping set : $$K(f) = \{ z \in \mathbb{C} : f^{(k)}(z) \nrightarrow_{k \to \infty} \infty \} $$

Question : If $K(f)$ is connected, is it also contractible ?

Let $f$ be a function, holomorphic in $\mathbb{C}$, and $K(f)$ its non-escaping set : $$K(f) = \{ z \in \mathbb{C} : f^{(k)}(z) \nrightarrow_{k \to \infty} \infty \} $$

Question : If $K(f)$ is connected, is it also contractible ?

Let $f$ be an holomorphic function and $K(f)$ its non-escaping set (also called the filled Julia set) : $$K(f) = \{ z \in \mathbb{C} : f^{(k)}(z) \nrightarrow_{k \to \infty} \infty \} $$

Question : If $K(f)$ is connected, is it also contractible ?

Let $f$ be an holomorphic function and $K(f)$ its non-escaping set : $$K(f) = \{ z \in \mathbb{C} : f^{(k)}(z) \nrightarrow_{k \to \infty} \infty \} $$

Question : If $K(f)$ is connected, is it also contractible ?