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



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> **Question** : If $K(f)$ is connected, is it also contractible ?
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Contractibility of connected holomorphic dynamics?
The answer to your question is negative. 


**EDIT** I have added some additional details and made some corrections.


**In the entire case**, it is possible to construct an entire function with the following properties: 
(a) The Fatou set consists of a single conneted attracting basin;
(b) If $C$ is a component of the Julia set, then the set of non-escaping points in $C$ is totally disconnected (in fact has Hausdorff dimension zero),
(c) There is a component of $J(f)$ that contains a non-escaping point, but no point that is accessible from $F(f)$.


Now, by (a), the nonescaping set is connected (since it contains a dense connected subset of the plane). On the other hand, it can be shown that there is no curve connecting the non-escaping points in (c) to a point in the Fatou set without intersecting the escaping set. 


Hence the non-escaping set is not path-wise connected, and hence not contractible. (The construction is contained in an upcoming article of mine, dealing more generally with the topology of transcendental Julia sets.) 


For **quadratic Cremer polynomials**, the key point is that the Cremer point $z\_0$ is accumulated on by small cycles by work of Yoccoz. Now, if the Julia set is path-connected (otherwise, there is nothing to prove), then there is a unique arc connecting each of these periodic points to $z\_0$. 


Now, for any cycle, it follows from the work of Perez-Marco that at least one of the corresponding arcs has diameter at least $\delta$, for $\delta$ independent of the cycle. Indeed, I believe it follows that at least one of them must contain the critical point. 


From this, one can deduce (although I haven't made sure to check all the details) that the Julia set is not contractible. 


