Can anybody give me a simple proof that simple covers of a Riemann surface have no covering automorphisms?

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    $\begingroup$ What's a simple cover? A one-sheeted covering space? $\endgroup$ – jef808 May 7 '15 at 17:01
  • $\begingroup$ it is a cover that has only simple branching, i.e. of type $z \mapsto z^2$ $\endgroup$ – IMeasy May 8 '15 at 9:16
  • $\begingroup$ The example you wrote is a counterexample: $z \mapsto -z$ is a covering automorphism of $z \mapsto z^2$. $\endgroup$ – Lee Mosher May 8 '15 at 14:16

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