Rational Maps don’t have Levy cycles

Here is a sketch of the following:

Theorem. The rational functions doesn’t have Levy cycles.

Proof. Suppose it does and denote by \Gamma the corresponding Levy cycle. Choose \gamma_1,\gamma_2,\ldots, \gamma_n the geodesics representatives in \Gamma. Then the map f: \bar{C}-f^{-1}(P_f) \to \bar{C}-P_f is the covering map, hence a local isometry and preserves the distances locally. Therefore

l_{\bar{C}-f^{-1}(P_f)} (\gamma'_i) = l_{\bar{C}-P_f} (\gamma_{i+1}) where \gamma'_i is the component of f^{-1}(\gamma_{i+1}) which is in the same homotpy class with \gamma_i. From the other hand \bar{C}-f^{-1}(P_f) \subset \bar{C}-P_f and

l_{\bar{C}-P_f} (\gamma_{i+1}) =  l_{\bar{C}-f^{-1}(P_f)} (\gamma'_i)  >  l_{\bar{C}-P_f} (\gamma_i) and we get that each \gamma_i is strictly shorter than \gamma_{i+1}, which is impossible.

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