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At a fixed point $y=x$, and the first equation factors. The branches areThe three branches meet at $a=0$, and the positive square-root branch meets the quadratic branch at $a=1$. The Jacobian matrix at a general fixed point isOn $x=a^2$, its trace is $-1$ and determinant $a-a^4=a(1-a^3)$: the branch is attracting for $0<a<1$ and a saddle for $a<0$ or $a>1$. On $x=\pm \sqrt{a}$, the eigenvalues are $-1$ and $2a(1\mp a^{3/2})$. The negative branch is always a saddle for $a>0$; the positive branch is a saddle for $0<a<1$ and attracting for $a>1$. At the bifurcation values the zero eigenvalue requires the nonlinear analysis below.

For $a=1+\mu$, $u=x-1$, $v=y-x$, the equations areThe extended centre manifold has $v=h(u,\mu )$ with no linear terms. Since $h_u\dot{u}$ has degree at least three, its invariance equation to degree two is $0=-h_2-(2{\mu }^2-5\mu u+2u^2)$. ThusThe reduced quadratic factors as $(u-2\mu )(2u-\mu )$. Its two roots have reduced eigenvalues $3\mu$ and $-3\mu$, respectively, and the transverse eigenvalue remains negative. They exchange stability at $\mu =0$, a transcritical bifurcation. The exact branches have $u=2\mu +{\mu }^2$ and $u=\mu /2-{\mu }^2/8+\cdots$, consistent with this second-order evolution equation.

Near $a=0$, the slow equation has leading form $\dot{x}=x^3-ax$, so this is a subcritical pitchfork bifurcation with the parameter orientation displayed. For $a<0$ there is one nearby saddle, whose unstable direction is close to $y=x$ and stable direction close to vertical. For $a>0$ the central branch is attracting and the outer two equilibria near $\pm \sqrt{a}$ are saddles; their separatrices distinguish attraction toward the central equilibrium from departure outwards. The actual central location is $(a^2,a^2)$, not exactly the origin away from $a=0$.

Adding positive $\epsilon$ breaks the pitchfork symmetry. The leading equilibrium equation near zero is $x^3-ax+\epsilon =0$: the diagram unfolds into a single branch plus a saddle-node pair, with the latter appearing for positive $a$ of order ${\epsilon }^{2/3}$. More precisely the fold is at $x=(\epsilon /2{)}^{1/3}$, $a=3(\epsilon /2{)}^{2/3}$ at leading order. Near one the reduced equation becomes $2u^2-5\mu u+2{\mu }^2+\epsilon =0$. Its discriminant is $9{\mu }^2-8\epsilon$; the crossing is replaced by two saddle-node endpoints with a no-equilibrium gap for $|\mu |<\sqrt{8\epsilon }/3$ to leading order. Outside that gap, the lower reduced root is stable and the upper one unstable. These local approximations describe the sketch; the exact perturbed curves have higher-order corrections.

Let the three orbit points be $a<b<c$. Up to reflection of the interval, their images are $F(a)=b,F(b)=c,F(c)=a$. Set $I=[a,b]$ and $J=[b,c]$. Continuity and the intermediate value theorem imply the covering relationsWe first prove the needed interval-covering principle using the permitted pullback fact. If a cyclic list of closed intervals $I_0,\dots ,I_{n-1},I_n=I_0$ obeys $F(I_j)\supseteq I_{j+1}$, pull back successively from the end to obtain a closed interval $K\subseteq I_0$ with $F^n(K)=I_0$ and $F^j(K)\subseteq I_j$. Since $F^n(K)\supseteq K$, the intermediate value theorem gives a fixed point of $F^n$ in $K$: choose preimages of the two endpoints of $K$, which give opposite weak signs of $F^n(x)-x$, irrespective of their order. This point realizes the cyclic itinerary.

For $n\ge 2$, use the word $I{J}^{n-1}$. It obeys the covering relations and therefore yields a point with period dividing $n$. A point away from the shared endpoint $b$ has an unambiguous itinerary, and this word has exactly one $I$ per cycle; it cannot be a repetition of a shorter word. Its least period is consequently $n$. If an orbit touches $b$, it is the original three-cycle. That realizes the case $n=3$ itself. It cannot realize any of the other words: for $n$ not divisible by three it is not fixed by $F^n$, and for a multiple of three larger than three it visits $a\in I\setminus J$ more than once, contradicting the itinerary. Thus the construction works for every $n\ge 2$. For $n=1$, $F(J)\supseteq J$ and the same fixed-point argument gives a fixed point. This proves period three implies all periods: there is an orbit of every positive least period.

For the two five-cycles forced by a three-cycle, use the two cyclic words $IJJJJ$ and $IJIJJ$. Both are admissible. Their points cannot be endpoints from the three-cycle, and because five is prime and the words use both intervals, each has least period five. Their itineraries have respectively one and two visits to the interior of $I$; cyclically shifting the starting point cannot change this count. Hence they give at least two distinct period-five orbits.

A fixed point $x_0$ is Lyapunov stable if for every $\epsilon >0$ there is $\delta >0$ such that solutions starting within $\delta$ of $x_0$ stay within $\epsilon$ for all forward times. It has quasi-asymptotic stability if some neighborhood of $x_0$ consists of initial points whose solutions converge to $x_0$. Asymptotic stability requires both properties.

Choose the Lyapunov functionIt vanishes only at the origin, is positive elsewhere, and has compact sublevel sets. The vector field is polynomial, hence locally Lipschitz. Along a solution,A small sublevel set is forward invariant and can be fitted inside any prescribed neighborhood, proving Lyapunov stability; boundedness also prevents finite-time escape.

For attraction, a bounded forward solution has a nonempty compact limit set, and $V$ decreases to a limit. Every point in that limit set has that same $V$ value. Continuous dependence of solutions makes the limit set forward invariant, so $\dot{V}=0$ along every trajectory in it. Such a trajectory must have $x=0$ at all times. But then $\dot{x}=-y$, forcing $y=0$. Thus the only possible limit point is the origin, and every solution converges there. This verifies the invariant-set condition behind the LaSalle invariance principle, rather than incorrectly treating $\dot{V}$ as strictly negative at every nonzero point. The origin is therefore asymptotically stable, indeed globally attracting.

For a nonconstant holomorphic map $f:X\to Y$ between compact connected Riemann surfaces, its degree of a holomorphic map is the number of preimages of a regular value. More generally it is the constant sum ${\sum }_{p\in f^{-1}(q)}{e}_p$, where $e_p$ is the local multiplicity. The Riemann-Hurwitz formula is

Negation is well-defined on the complex torus because $-\Lambda =\Lambda$. It is holomorphic with holomorphic inverse itself, so $\psi$ is biholomorphic. Its fixed points satisfy $2z\in \Lambda$. If $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$, they are the four distinct classes

Away from these points, choose a small coordinate disk $U$ disjoint from $\psi (U)$. The quotient map is a homeomorphism from $U$ onto $\pi (U)$, so transfer its complex coordinate to that open set of $S^{\prime }$. The transition between two such lifted charts is induced either by identity or by $\psi$, hence is holomorphic. These charts give the required Riemann-surface structure, and $\pi$ is locally a holomorphic bijection there.

The quotient has generic degree two. At a fixed point, a local coordinate centred there changes under $\psi$ as $u\mapsto -u$. The assumed holomorphic quotient map is invariant under this operation, so its local expansion is even and has multiplicity at least two. Since the total degree is two, it is exactly two. There are no other ramification points. The domain torus has genus one, so Riemann-Hurwitz formula gives

For $t>0$, the three-dimensional Kirchhoff formula with zero displacement isThe Strong Huygens principle says that the value depends only on the initial velocity on the sphere of radius $t$, not on points strictly inside it. In particular a compactly supported disturbance leaves no tail at a fixed point after the sphere has passed the support.

For method of descent for the wave equation, extend the two-dimensional initial velocity independently of the third coordinate. Projecting the three-dimensional sphere onto its planar disk gives two sheets, each with area element $tdy/\sqrt{t^2-|y-x{|}^2}$. ThereforeThe local argument can use compact cutoffs in the extra coordinate, so global Schwartz decay there is unnecessary. The weak Huygens principle is finite propagation speed: only data in the closed disk of radius $t$ can influence the point. Unlike the three-dimensional free equation, the full interior contributes and a tail can persist.

For the equation with potential $|x{|}^2$, multiply by $u_t$ and use the product rule:This is the claimed local energy conservation with flux $p=-u_t\nabla u$.

Fix $(t_0,x_0)$ and let $E(t)$ be the integral of $e$ over the shrinking ball $|x-x_0|\le t_0-t$, $0\le t\le t_0-a$ with $a>0$. Its boundary moves inward at speed one. The divergence theorem givesbecause the negative integrand isAll initial data are zero, so $E(0)=0$. Nonnegativity then forces $E(t)=0$. Letting the small terminal ball approach the cone tip proves $u=0$ there; alternatively $u_t=0$ throughout the cone and the initial value fixes $u$. The same argument applied to a difference of solutions whose data agree on the initial ball proves local domain of dependence. Thus the potential equation has propagation speed at most one and satisfies the weak Huygens principle, without requiring finite total energy on all space. It does not assert the strong no-tail property.

The fixed points are $(0,0)$ and $(\pm 1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, so it is a saddle with stable direction the y-axis. At either other fixed point, the characteristic polynomial is ${\lambda }^2+\lambda +2$, giving eigenvalues $(-1\pm i\sqrt{7})/2$. Thus both are stable foci. The nullclines are $x=0$, $y=1$ and $y=x^2$. In the right half-plane $x$ grows below $y=1$ and decreases above it; in the left half-plane these horizontal arrows reverse. Vertically, $y$ increases below $y=x^2$ and decreases above it.

A global argument determines the requested omega-limit sets, rather than guessing from the local phase portrait. For $x\ne 0$, use the quadratic-logarithmic Lyapunov function for a feedback systemIts sublevel sets are compact and bounded away from $x=0$, so trajectories starting in either half-plane exist for all positive time and stay in such a set. Since $L$ decreases, it tends to a limit, and is constant on the omega-limit set. Every point of that set therefore has $y=1$. Invariance then requires $\dot{y}=x^2-1=0$, so only the corresponding focus can occur. Also $x(t)=x(0)\exp ({\int }_0^t(1-y(s))ds)$ preserves its sign.

ConsequentlyOn $x=0$, the exact solution is $y(t)=y(0)e^{-t}$. The phase portrait below shows this stable axis, the nullclines and the spiral attraction in each half-plane.

For the given Möbius transformation, the fixed point equation is $z(z+3)=3z+1$, hence

For the general Möbius transformation, its finite fixed points satisfyWith $c\ne 0$ and nonzero discriminant, the quadratic formula gives the two distinct fixed points of a Möbius transformation:The fixed-point equation for $\alpha$ also gives $b-d\alpha =-\alpha (a-c\alpha )$. ThereforeDividing yields the conjugation of a Möbius transformation with two distinct fixed points:Neither factor in this quotient vanishes, since $(a-c\alpha )(a-c\beta )=ad-bc\ne 0$. The formula extends over the Riemann sphere, including the transformation's pole. For the special transformation and the ordering $\alpha =1,\beta =-1$, $k=1/2$.

At a fixed point $x=y=p$, the quadratic equation is $p^2+(1+b)p-\mu =0$. ConsequentlyThe Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -b& -2p\end{array}\right)$, and its Floquet multipliers solve ${\lambda }^2+2p\lambda +b=0$. The Jury stability criterion requires $|b|<1$ and $1+2p+b>0$, $1-2p+b>0$. With $b>-1$, this becomes $-1<b<1$, $-(1+b)/2<p<(1+b)/2$. The lower branch never satisfies it; the upper branch is attracting precisely whenAt the lower boundary the two fixed points merge with Floquet multipliers $1,b$, a saddle-node bifurcation for $b\ne 1$. At the upper boundary the upper point has Floquet multipliers $-1,-b$, giving a generic period-doubling bifurcation for $-1<b<1$.

Its direction can be established explicitly. A distinct two-cycle alternates values $y,z$ satisfying $(1+b)z=\mu -y^2$, $(1+b)y=\mu -z^2$. Subtracting gives $y+z=1+b$, and substitution givesIt exists beyond the fixed-point period-doubling bifurcation. The second-iterate Jacobian matrix has determinant $b^2$ and trace $4yz-2b=4[(1+b{)}^2-\mu ]-2b$. Applying the same stability inequalities shows attraction for $-1<b<1$ andThus the first period-doubling bifurcation is supercritical; the next upper boundary has a Floquet multiplier $-1$ for the two-cycle. These are the fixed-point and two-cycle thresholds of the Hénon map.

At $b=1$ the map preserves phase-space area, so attraction is impossible. The fixed-point fold is at $\mu =-1$, with double Floquet multiplier $+1$, and is a conservative saddle-centre bifurcation. The upper branch $p_{+}=-1+\sqrt{1+\mu }$ is linearly elliptic for $-1<\mu <3$; the lower branch is a saddle fixed point of a map. At $\mu =3$, the upper branch has double Floquet multiplier $-1$ and becomes a saddle fixed point of a map for larger $\mu$. The two-cycle $y,z=1\pm \sqrt{\mu -3}$ is elliptic for $3<\mu <4$, then reaches double Floquet multiplier $-1$ for the second iterate at $\mu =4$. The partial sketch distinguishes elliptic, neutrally stable Floquet multipliers from attracting states. Resonances may require additional nonlinear analysis; unit Floquet multiplier moduli alone do not prove nonlinear Lyapunov stability.

The $b=1$ segment is a conservative Hénon stability boundary, not an ordinary nondegenerate dissipative Neimark–Sacker bifurcation. Indeed for $b>0$ constant phase-space area scaling is $\mathrm{area}(F(D))=b\mathrm{area}(D)$. A simple invariant closed curve would enclose a bounded invariant set $D$ of positive phase-space area, forcing $b=1$. Thus no such closed invariant curve can bifurcate into $b\ne 1$ within this constant-determinant family.

Generic folds and flips away from these degeneracies persist under sufficiently small smooth perturbations: their simple unit Floquet multiplier, transverse parameter crossing and nonzero leading nonlinear coefficient persist, while the other Floquet multiplier stays away from the unit circle. The implicit function theorem then continues the bifurcation curves and their criticality. The unqualified perturbation claim needs this restriction. Arbitrary perturbations need not preserve the conservative $b=1$ behavior. For example, in local complex coordinates at a nonresonant elliptic fixed point, the area-preserving map property gives a cubic normal form $z^{\prime }=e^{i\theta }z(1+i\tau |z{|}^2)+\cdots$ with zero cubic radial damping. Composing with the arbitrarily small radial perturbation $z\mapsto (1-\eta |z{|}^2)z$ introduces radial coefficient $-\eta$. Varying the modulus through one can now give a genuine Neimark–Sacker bifurcation. Conservative degeneracies and resonances are therefore not covered by a blanket structural-stability assertion.

Write $k=\sqrt{\mu }$. The equilibrium points are a stable node $(-k,0)$ and a saddle equilibrium $(k,0)$, whose unstable and stable eigenvalues are $2k$ and $-\lambda$. The saddle equilibrium's local stable manifold is $x=k$. When the right-going unstable separatrix returns at $(\nu ,h)$ with $\nu =k$, it lands exactly on that stable manifold, producing a homoclinic orbit of the saddle. Crossing this condition changes whether the returning trajectory is captured by the node or lies on the outgoing side of the saddle equilibrium.

Use an incoming section $y=h$ and outgoing section $x=h$. For entry $(x,h)$ with $k<x<h$, integrate the local flow:so the outgoing height isThe global excursion is smooth and transverse. Its return coordinate has expansion $x_{\mathrm{next}}=\nu +cY+O(Y^2)$, with $c>0$ in these section orientations. Thus the local passage map near a dissipative saddle-node givesFor small $\mu >0$, $\lambda >2k$, hence $P(k^{+})=\nu$ and $P^{\prime }(k^{+})=0$. If $\nu -k>0$ is small, the graph starts above the diagonal and, with slope less than one nearby, crosses it close to $x=\nu$. More explicitly, $P(x)-x=(\nu -k)-(x-k)+O((x-k{)}^{\beta })$, giving a unique small fixed point with Floquet multiplier $0<P^{\prime }<1$. It corresponds to an attracting periodic orbit. For $\nu <k$ no such local crossing exists on $x>k$; the unstable branch returns to the node's side instead. The Poincaré return map panel shows this geometric fixed-point test.

For $\mu <0$, put $k=\sqrt{-\mu }$. There are no local equilibrium points and $\dot{x}=x^2+k^2>0$, so every incoming point passes through the local region. The same sections giveand again $P(x)=\nu +cY(x)+O(Y^2)$. The global return supplied by the problem closes the passage into a periodic orbit if this map has a fixed point.

Its derivative is strongly contracting: $Y^{\prime }(x)=\lambda Y(x)/(x^2+k^2)$. To see the smallness uniformly near the codimension-two point, put $d=\sqrt{x^2+k^2}$ and assume $2d<h$. The passage includes $[d,2d]$, on which ${\xi }^2+k^2\le 5d^2$, so $T_{\mathrm{loc}}\ge 1/(5d)$. ConsequentlyFor a sufficiently small entry interval and all sufficiently small $\nu$ of either sign, $P$ maps the interval into itself and has derivative bounded strictly below one. The mean value theorem bounds differences by a factor strictly below one, while the endpoint signs of $P(x)-x$ give existence of a fixed point. ThusHere “all $\nu$” means all return parameters in the neighborhood where the stated local/global construction applies, not arbitrarily distant global return geometries.

For $\mu >0$, the left equilibrium point $(-\sqrt{\mu },0)$ is a stable node, the right equilibrium point $(\sqrt{\mu },0)$ a saddle equilibrium, and the saddle equilibrium's left unstable branch flows directly towards the node. The right branch takes the prescribed global excursion. Its landing relative to the two equilibrium points gives the six local portraits in the figure.

If $\mu >{\nu }^2$, then $-\sqrt{\mu }<\nu <\sqrt{\mu }$ for either sign of $\nu$. The global branch lands between the equilibrium points and flows left into the node. There is no nearby Poincaré return map cycle, and the landing approaches the node from its right.

If $0<\mu <{\nu }^2$ and $\nu >0$, the branch lands to the right of the saddle equilibrium. The contracting Poincaré return map has an attracting fixed point, giving a stable periodic orbit; the node also remains as a separate attractor with its own basin. If instead $\nu <0$, the branch lands to the left of the node and approaches it from the left. There is no nearby periodic orbit in that case.

If $\mu <0$, the two equilibrium points have vanished and a stable returning cycle exists for either sign of $\nu$. For negative $\nu$ the cycle traverses the full bottleneck and has the divergent period in part (c). For fixed positive $\nu$ it enters to the right of the bottleneck; its local period approaches $1/\nu -1/h$, so it need not diverge when $\mu \to 0^{-}$. The outer arcs in the figure encode only the given return; the local equations do not specify the full vector field away from the origin.

The boundaries explain the codimension-two organization. On $\nu =\sqrt{\mu }>0$ the saddle equilibrium has a homoclinic orbit. On $\mu =0$, $\nu <0$, the return passes through the semistable equilibrium point at a saddle-node bifurcation, producing a saddle-node bifurcation on an invariant circle. On $\mu =0$, $\nu >0$, the node and saddle equilibrium instead collide away from the persistent cycle. At $\mu =\nu =0$, the returning separatrix hits the equilibrium point's strong stable manifold at the saddle-node bifurcation: this is the saddle-node separatrix-loop bifurcation. The curve $\nu =-\sqrt{\mu }$ changes which side of the node receives the returning orbit; unlike the positive branch, it is not a global bifurcation through a saddle homoclinic orbit.

Let $R$ be a nonconstant rational map of the Riemann sphere, and write $R^n$ for its $n$-fold iterate, with $R^0$ the identity. The appropriate notion of convergence is uniform convergence on compact subsets in the chordal metric, which treats poles as ordinary values at infinity. A family of meromorphic maps is a normal family if every sequence in it has a subsequence converging locally uniformly in that metric; the limit is meromorphic or constant, possibly the constant infinity.

The Fatou set $F(R)$ consists of points with a neighborhood on which the iterate family is normal. The complementary Julia set is $J(R)=\hat{\mathbb{C}}\setminus F(R)$. Thus the sphere is the disjoint union of an open stability region $F(R)$ and a closed instability region $J(R)$. These definitions concern all subsequences of iterates, rather than assuming that the iterates themselves converge on every stable region.

Equicontinuity makes this distinction precise. On a compact subset $K$ of a neighborhood, the family is equicontinuous if, for every $ϵ>0$, some $\delta >0$ makes $\chi (R^n(z),R^n(w))<ϵ$ whenever $z,w\in K$ and $\chi (z,w)<\delta$, uniformly in $n$. The Arzelà-Ascoli theorem says that equicontinuity of continuous maps from a compact domain to a compact metric target gives relative compactness for uniform convergence. Apply it to a compact exhaustion and take a diagonal subsequence to obtain local normality. Locally uniform spherical limits of meromorphic maps are again meromorphic or constant, by using finite target charts and the locally uniform limit theorem for holomorphic functions. Conversely, if equicontinuity failed, there would be maps $f_j$ and points $z_j,w_j\to z$ with $\chi (f_j(z_j),f_j(w_j))$ bounded below by a positive number. A subsequence converging uniformly on a compact neighborhood contradicts this. Hence local normality and local equicontinuity are equivalent here.

Both sets are completely invariant:For the inverse direction, if the family is normal near $R(z)$, composition with $R$ makes $R^{n+1}$ normal near $z$, and adding the identity does not affect normality. For the forward direction, suppose the iterates are equicontinuous near $z$. A nonconstant holomorphic sphere map is open. Thus each sufficiently small neighborhood of $z$ maps onto a neighborhood of $R(z)$; every nearby target point $w$ has a preimage $x$ as close to $z$ as desired. Thenis uniformly small. Applying this at each point of the image neighborhood proves equicontinuity there. This argument works at rational critical points and requires no single-valued inverse branch. Surjectivity of $R$ supplies the asserted image equalities. Also $F(R^k)=F(R)$ and $J(R^k)=J(R)$ for $k\ge 1$: one direction follows by taking a subfamily, and the other by splitting $R^n$ into the finitely many residue classes $R^r\circ (R^k{)}^m$, $0\le r<k$.

For degree $d\ge 2$, the Julia set is nonempty. Otherwise the family would be equicontinuous on the whole compact sphere, giving a subsequence $R^{n_j}$ converging uniformly to a continuous sphere map $G$. Maps sufficiently close in chordal distance are homotopic, using the unique short spherical geodesic between their values. They therefore have equal topological degrees. This is impossible because $\mathrm{deg}{R}^{n_j}=d^{n_j}\to \infty$ whereas $G$ has one fixed degree.

It cannot be a nonempty finite set either. Complete invariance would make $R$ a permutation of that finite set, with just one sphere preimage for each of its points. That preimage would have local degree of a holomorphic map $d$. Every holomorphic cycle in this set would consequently be superattracting for its return map, hence lie in $F(R)$, a contradiction. In particular $J(R)$ contains at least three points.

The Montel theorem in the form used here says that a family of meromorphic maps on a domain, each omitting the same three distinct points of the sphere, is normal for spherical convergence. It also shows that $J(R)$ is perfect. If $a$ were isolated in $J(R)$, choose a disc $U$ with $U\cap J(R)=\{a\}$ and four distinct points of $J(R)$. For each $n$, complete invariance makes $R^n(U)\cap J(R)$ have at most the single point $R^n(a)$. Partition the iterates into finitely many families according to which of the four points, if any, that single value equals. Each family omits at least three fixed points, so each is normal by Montel; their finite union is normal. This would put $a$ in $F(R)$, a contradiction. Therefore the degree-at-least-two Julia set is compact, infinite and has no isolated points.

The same omitted-value argument explains its global reach. If an open $U$ meets $J(R)$, then ${\bigcup }_{n\ge 0}{R}^n(U)$ can omit at most two sphere points, since three omissions would imply normality on $U$. If $J(R)$ had an interior and were not the whole sphere, these image sets would remain in $J(R)$ and their complement would contain the nonempty open set $F(R)$, impossible. Thus a proper Julia set has empty interior and equals the boundary of $F(R)$. If the Julia set is the whole sphere, the Fatou set is empty instead.

An attracting periodic point and its holomorphic cycle has a contracting return map in a small disc, so its basin is in the Fatou set. Repelling periodic points belong to the Julia set: return iterates fix the point but have derivatives equal to unbounded powers of their periodic-orbit multiplier, contradicting derivative bounds for a normal subsequence near that fixed value. A standard further global property is that repelling periodic points are dense in the Julia set. Fatou components can also support convergence towards a parabolic cycle on the boundary, or rotation behavior in a Siegel disc or a Herman ring; normality need not mean attraction. For example, $R(z)=z^d$ has $J(R)=\{|z|=1\}$, with the interior converging to zero and the exterior to infinity. On the unit circle, arbitrarily close radii have incompatible limiting behavior, preventing normality there.

Degree one has a different picture. A rational map of degree one is a Möbius transformation. Up to Möbius conjugacy, an elliptic map is $z\mapsto az$ with $|a|=1$; its iterates are rotations and are equicontinuous on the sphere, so $F$ is the whole sphere and $J$ is empty (including the identity). For $z\mapsto az$ with $|a|>1$, all compact subsets away from zero converge to infinity, while $z_n=a^{-n}\to 0$ and $R^n(z_n)=1$ prevent equicontinuity at zero. Thus $J=\{0\}$; for $|a|<1$ the reciprocal coordinate gives $J=\{\infty \}$. A nonidentity parabolic map is conjugate to $z\mapsto z+1$. Its iterates converge locally uniformly to infinity on the finite plane, but $z_n=-n\to \infty$ has $R^n(z_n)=0$, so $J=\{\infty \}$. Holomorphic conjugacy carries these sets to the corresponding fixed points of the original map. These finite Julia sets explain why the degree restriction in many global Julia-set properties is essential.

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