I'm searching for a rigorous proof that $c$ is in the Mandelbrot set $M$ if and only if its Julia set $J_c$ is connected. I've read this question and answer , I understand the answer that is given however, first, I don't know how to prove the assertion he mentioned :
"the preimage of a curve will have one component if and only if $\ 0 \ $ lies in the the interior of the preimage. This is because $\ 0 \ $ is the critical point of the map $P_c$."
What I know is that the image of a connected set by a continuous function is a connected set. Furthermore, I would like to understand that if we assume that a Julia set is connected if and only if it contains $0$, how does it prove that $c \in M \Leftrightarrow J_c$ is connected ?
One can show that if $0$ $D \in J_f$, then $J_f$ is connected. In particular, let $U$ be some neighborhood of $\infty$, and note that $\mathbb C \setminus K$ is the "stuff that goes to infinity, or in other words:
$$\bigcup_{n=1}^{\infty} f^{-n}(U)$$ so that the filled julia set is $J_f:=\bigcap_{n=1}^{\infty} f^{-n}(U^c)$.
Note that $U^c$ is a topological disk, $D$.
question: when is $f^{-1}(D)$ a single disk (aka connected component)?
answer: when $D$ contains $c$ (or equivalently, $f^{-1}(D)$ contains $0$).
Well for $f^{\prime}(z)=2z=0$ when $z$ is zero, and there are two disks if the forward image does not contain $c$ (apply the riemann-hurwitz formula, where $\chi(D)=1$, but $z^2+c$ would be a double cover, so the characteristic of the preimage will be $2$.)
So, the intersection is connected if and only if the filled julia set contains zero. So, the values of $c \in \mathbb C$ where $0$ is bounded under $z^2+c$ are precisely the ones with connected julia sets.
**note: the image of a connected set is connected, but there is no theorem that says that the forward image of a disconnected set is not connected.