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A critical orbit is the forward sequence $p,R(p),R^2(p),\dots$ starting at a critical point of a rational map. The postcritical set records the forward images, usually from the first iterate onward, of all critical points; a single critical orbit and the union of all such images are distinct notions. For quadratic polynomials there is one finite critical point, zero, and its orbit controls connectedness of the filled Julia set and membership in the Mandelbrot set. A bounded orbit can be preperiodic to a repelling cycle, as $0\mapsto i\mapsto -1+i\mapsto -i\mapsto -1+i$ for $z^2+i$; boundedness then does not imply Fatou membership.

For quadratic-polynomial parameter space, these are open components of the interior of the Mandelbrot set with an attracting periodic orbit.

A hyperbolic component in a parameter space consists of maps with a specified attracting cycle. Its multiplier gives a holomorphic coordinate on the component in the quadratic family, and root-of-unity boundary multipliers give parabolic parameters.

The main cardioid consists of quadratic parameters with an attracting fixed point. If that fixed point has multiplier $\lambda$, then it is $\lambda /2$ and $c=\lambda /2-{\lambda }^2/4$, $|\lambda |<1$. The map is injective on the open unit disc: equality of two parameter values forces either equal multipliers or their sum to be two, impossible for distinct interior multipliers. Its boundary is the cardioid obtained from $\lambda =e^{it}$. At $\lambda =1$ the parameter derivative vanishes, giving the cusp $c=1/4$. Parameters just to the real right of that cusp have monotonically escaping critical orbits, while the cusp's critical orbit tends to $1/2$, so it is on the Mandelbrot set boundary.

For $f_c(z)=z^2+c$, the Mandelbrot set is the set of parameters for which the finite critical orbit $0,f_c(0),f_c^2(0),\dots$ 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.

The affine map $h(z)=1/2-z$, its own inverse, givesThis 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 $c=1/4$ 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 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 $w=1/z$,For $|w|\le 1/4$, its modulus is at most $|w|/3$. Thus the reciprocal iterates tend uniformly to zero there, proving 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 becomesIf $\Re w>1$, then $\Re (1/(w-1))>0$, so $\Re T(w)>\Re w+1$. Consequently the half-plane is forward invariant andThis 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 $P^n\to 0$ 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 $U$. ThereforeThis 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. Butfor every $n$, whereas the derivative of the zero limit is zero. This contradiction proves

Write $n=km$, with positive integers $k,m$, and put $Q=P^m$. Polynomial substitution respects congruence, so $Q(z)\equiv z(\mathrm{mod}Q(z)-z)$ implies inductively $Q^j(z)\equiv z$ for every $j\ge 1$. Since $Q^k=P^n$,Equivalently one can factor each $Q^{j+1}(z)-Q^j(z)$ by $Q(z)-z$ and telescope. If $Q(z)-z$ is identically zero, then $Q$ is the identity and $P^n-z$ is also zero, so the assertion still holds in the sense that the latter is a polynomial multiple of the former.

For the holomorphic cycle, the chain rule giveswith indices cyclically understood. These are the same product at every point. If its modulus is less than one at one point, it is less than one at every point. Taylor expansion of each return map gives a contracting neighborhood of its fixed point, exactly as in question 1. Hence every point of the holomorphic cycle is attracting for $P^n$ whenever one is; a zero factor makes every return periodic-orbit multiplier zero.

For the quadratic family, factor the second-iterate fixed-point equation:The first factor gives fixed points. A root of the second is also fixed only if subtracting the two factors gives $2z+1=0$, so $z=-1/2$ and $c=-3/4$. If $c\ne -3/4$, the second factor has discriminant $-4c-3\ne 0$ and has two distinct roots, neither fixed. They are exchanged by $P$, since each is fixed by $P^2$ but not by $P$, and form an exact two-cycle. Therefore the unique parameter with no exact two-cycle isAt this parameter,Thus the four finite solutions counted with multiplicity are $3/2$ once and $-1/2$ three times: there are only two distinct finite solutions, both already fixed by $P$. The periodic-orbit multiplier at $3/2$ is three (nine for $P^2$), so it is repelling. At $-1/2$ the periodic-orbit multiplier for $P$ is minus one; for $P^2$ it is one. Indeed, with $u=z+1/2$, the local first-return expression for $P$ is $-u+u^2$, and its square is $u-2u^3+u^4$. This is a parabolic return with a triple fixed-point root. The two-cycle has merged into this fixed point; the triple root is not three distinct periodic points. In the Mandelbrot set picture this is the period-two bulb's attachment to the main cardioid.