 Completely invariant set of a rational map

A subset $E$ is completely invariant for a rational map if $R(E)=E=R^{-1}(E)$. This asserts invariance in both directions, including every preimage, rather than just invariance of one chosen inverse branch. For a nonconstant sphere map, the two inclusions $R(E)\subseteq E$ and $R^{-1}(E)\subseteq E$ already imply the equalities, because the map is surjective. The Fatou set and Julia set are completely invariant. A Fatou component need not itself be completely invariant even though the union of all components is.