= Fatou component containing an entire fiber is completely invariant
{title2=$R(U)\subseteq U,\ R^{-1}(a)\subseteq U\Longrightarrow R^{-1}(U)=U$}
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.
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