Fatou component ($U\subseteq F(R)$)

A Fatou component is a connected component of the Fatou set of a rational map. It is open. A rational map sends each Fatou component onto another one: complete invariance of the Fatou set puts it inside the inverse image of the target component; its maximal connectedness makes it one full inverse-image component. Restricting the globally proper sphere map to such a component is proper, since the inverse image of a compact subset intersected with the component is closed in a compact set. Its image is both open and closed in the connected target component, hence the whole target. A component is periodic if some iterate returns it to itself.

A Herman ring is a doubly connected periodic Fatou component on which the first-return map is holomorphically conjugate to an irrational rotation of an annulus $r<|w|<1$. Compactness of the circle of rotation factors proves normality of the return iterates, and the finitely many intervening iterates preserve normality. It is the annular rotation-domain counterpart of a Siegel disc.

A Siegel disc is a simply connected periodic Fatou component on which the first-return map is holomorphically conjugate to an irrational rotation of the unit disc: $h\circ R^p\circ h^{-1}(w)=e^{2\pi i\theta }w$, with $\theta \notin \mathbb{Q}$. Its return iterates are normal because any sequence of rotation factors has a convergent subsequence in the unit circle, giving locally uniform limits after conjugation. The iterates generally do not converge to a constant. This is a rotation domain, illustrating stability without attraction.

Suppose a Fatou component $U$ satisfies $R(U)\subseteq U$, and some $a\in U$ has $R^{-1}(a)\subseteq U$. Every connected component of $R^{-1}(U)$ maps properly and surjectively to $U$: properness follows by intersecting compact preimages with the relatively closed component; openness and closedness then give surjectivity. Such a component must meet the entire fiber over $a$, which lies in $U$. Since $U$ is itself one inverse-image component, there can be no other. Hence $R^{-1}(U)=U$ and $R(U)=U$, proving complete invariance.