## 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.