= Solution
For $f_c(z)=z^2+c$, the <Mandelbrot set> is the set of parameters for which the finite <critical orbit> $0,f_c(0),f_c^2(0),\ldots$ remains bounded. It is a <compact> connected parameter set. Its <main cardioid> consists of parameters with an attracting <fixed point>, with a <period-two bulb> attached on the left, smaller bulbs attached recursively and fine filamentary structure. It is symmetric about the real axis and meets that axis in $[-2,1/4]$. The quadratic <filled Julia set> is connected exactly when $c$ belongs to this parameter set; outside it, the <rational critical point> escapes and the <Julia set> is a <Cantor set>. Thus it organizes the contrasting phase-space dynamics of the quadratic family, rather than itself being a <Julia set> in the dynamical $z$-plane.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-9-mandelbrot.png]
{title=Informal Mandelbrot-set sketch, with the cardioid cusp and period-two attachment marked}
The affine map $h(z)=1/2-z$, its own inverse, gives
$$
\boxed{h\circ P\circ h^{-1}(w)=w^2+1/4.}
$$
This value is unique even among <Möbius conjugacies>. The finite <fixed point> of $P$ is zero and its <periodic-orbit multiplier> is one. <Holomorphic conjugacy> preserves that <periodic-orbit multiplier>. Infinity has <periodic-orbit multiplier> zero for a quadratic <polynomial>, so a corresponding multiplier-one <fixed point> of $w^2+c$ must be finite; it obeys $2w=1$ and $w^2+c=w$, forcing $c=1/4$. To locate it explicitly, a quadratic <fixed point> of <periodic-orbit multiplier> $\lambda$ gives $c=\lambda/2-\lambda^2/4$. The boundary parametrization $|\lambda|=1$ has its cusp at $\lambda=1$, so \b[$c=1/4$ is the right-hand cusp on the Mandelbrot boundary]. It is in the set: the <critical orbit> starts at zero, stays in $[0,1/2]$ and increases to the <fixed point> $1/2$. For real $c>1/4$, the increment $f_c(x)-x=(x-1/2)^2+c-1/4$ is strictly positive; a bounded <critical orbit> would have a limit solving $x^2+c=x$, which has no real solution. These nearby real parameters escape, proving that the cusp belongs to the boundary of the full <Mandelbrot set> as well.
The <fixed points> of $P$ on the sphere are zero and infinity. There is \b[no neighborhood of zero on which all iterates converge to zero], even pointwise: a negative real initial point satisfies $P(x)=x-x^2<x<0$, so its iterates decrease without a finite limit and escape to infinity. Every neighborhood of zero contains such points. At infinity there is a uniformly attracting neighborhood in the <chordal metric>. In the reciprocal coordinate $w=1/z$,
$$
\frac1{P(1/w)}=\frac{w^2}{w-1}.
$$
For $|w|\leq1/4$, its modulus is at most $|w|/3$. Thus the reciprocal iterates tend uniformly to zero there, proving \b[<uniform convergence> to infinity on the corresponding sphere neighborhood].
For the given disc, $|z-1/2|<1/2$ is equivalent to $|z|^2<\Re z$, and hence to $\Re(1/z)>1$. Under $w=1/z$ the iteration becomes
$$
T(w)=\frac{w^2}{w-1}=w+1+\frac1{w-1}.
$$
If $\Re w>1$, then $\Re(1/(w-1))>0$, so $\Re T(w)>\Re w+1$. Consequently the half-plane is forward invariant and
$$
\boxed{P(\Delta)\subseteq\Delta,\qquad |P^n(z)|<\frac1{n+1}\quad(z\in\Delta).}
$$
This proves <uniform convergence> to zero on the disc and therefore $\Delta\subset F(P)$. It is the explicit <parabolic disk for z minus z squared>.
Let $U=F_\Delta$. The iterate family is normal on the connected <open set> $U$. Every subsequential limit along $n_j\to\infty$ equals zero on $\Delta$; by the <identity theorem> it is zero throughout $U$. If the full sequence failed to converge uniformly on some <compact> subset, a further normal <subsequence> would contradict that conclusion. Hence \b[$P^n\to0$ locally uniformly throughout $U$]. Complete invariance of the <Fatou set> puts $P(U)$ in a <Fatou component>. Since $P(\Delta)\subset\Delta$, that component is $U$, so $P(U)\subseteq U$.
We prove the reverse-image assertion, rather than assuming that a basin has only one component. Each <connected component> $V$ of $P^{-1}(U)$ maps properly to $U$: for <compact> $K\subset U$, the sphere preimage $P^{-1}(K)$ is <compact>, and its intersection with $V$ is closed in it because $V$ is a relatively closed component of $P^{-1}(U)$. Its image is open by holomorphic openness. To prove it is closed in $U$, lift a convergent sequence of image points to preimages in $V$. Their images eventually lie in a <compact> neighborhood inside $U$; properness gives a convergent preimage <subsequence> with limit in $V$ and the desired limiting image. Thus the image is also closed, and equals the connected target $U$. Moreover $U$ itself is one such component: it lies in $P^{-1}(U)$ by forward invariance, and any connected inverse-image component containing it lies in $F(P)$ and hence in $U$ by maximality.
Now $1/4\in\Delta$ and its entire preimage is $\{1/2\}$, counted twice. Every component $V$ of $P^{-1}(U)$ must contain a preimage of $1/4$, so every one must be $U$. Therefore
$$
\boxed{P^{-1}(U)=U=P(U).}
$$
This is the <Fatou component containing an entire fiber is completely invariant> argument. The convergence on this component is locally uniform: it cannot be uniform on all of $U$, because $P^n(U)=U$ for every $n$ and $U$ contains nonzero points.
Finally suppose zero belonged to the <Fatou set>. A normal <subsequence> on a disc around zero would have limit zero, since that disc meets $\Delta$ in an <open set> where the entire sequence tends to zero. The <identity theorem> forces this limit to be zero on the whole disc. <Uniform convergence> in a smaller finite coordinate disc would then imply <derivative> convergence at zero by the <Cauchy integral formula>. But
$$
(P^n)'(0)=P'(0)^n=1
$$
for every $n$, whereas the <derivative> of the zero limit is zero. This contradiction proves
$$
\boxed{0\in J(P).}
$$
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