= Solution
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 $J(R)=\widehat{\mathbb C}\setminus F(R)$. Thus \b[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 $\epsilon>0$, some $\delta>0$ makes $\chi(R^n(z),R^n(w))<\epsilon$ 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 \b[local normality and local <equicontinuity> are equivalent here].
Both sets are completely invariant:
$$
\boxed{R^{-1}(F)=F=R(F),\qquad R^{-1}(J)=J=R(J).}
$$
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. Then
$$
\chi(R^n(w),R^n(R(z)))=\chi(R^{n+1}(x),R^{n+1}(z))
$$
is 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 $k\geq1$: one direction follows by taking a subfamily, and the other by splitting $R^n$ into the finitely many residue classes $R^r\circ(R^k)^m$, $0\leq r<k$.
For degree $d\geq2$, 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 $\deg R^{n_j}=d^{n_j}\to\infty$ 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 \b[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 $\bigcup_{n\geq0}R^n(U)$ 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 \b[$F$ 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 $z_n=a^{-n}\to0$ and $R^n(z_n)=1$ prevent <equicontinuity> at zero. Thus \b[$J=\{0\}$]; for $|a|<1$ the reciprocal coordinate gives \b[$J=\{\infty\}$]. 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 \b[$J=\{\infty\}$]. <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.
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