Fatou component containing an entire fiber is completely invariant ($R(U)\subseteq U,R^{-1}(a)\subseteq U⟹R^{-1}(U)=U$)

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.