If I have an automorphism of the open unit disc, then I think that it naturally extends to a holomorphic function from the closed unit disc into itself (use continuity to define values on the boundary), but I can't see why this new holomorphic function would map the unit circle onto itself? Why would points of the unit circle will map to points of the unit circle (continuity arguments?), and, more importantly, why would the image of the unit circle be all of the unit circle? I can't prove it using open mapping theorem/ other elementary arguments that come to mind. Please do not quote the automorphisms of the unit disc: I already know them, and this question actually sprung from motivating arguments for the theorem that states all automorphisms of the unit disc.

  • $\begingroup$ Actually none of the assertions in your question are obvious and all need some careful argument involving both analysis and topology. But fortunately, automorphism of unit disc can be written out explicitly, so the answer to your question would be easy. $\endgroup$ – Cave Johnson Dec 11 '17 at 4:23
  • $\begingroup$ Thank you for your comment. This question actually sprung from some remarks on page 182 of Bak and Newman which appear before they explicitly state the automorphisms of the unit disc: ''If we assume that $f$ is bilinear then since f is globally 1-1, it must map the unit circle onto itself...". This is why I've been puzzled for a while. $\endgroup$ – Acton Dec 11 '17 at 4:38
  • $\begingroup$ So you extend by continuity. How then do you know the resulting continuous function is holomorphic at points on the boundary. $\endgroup$ – Michael Hardy Dec 11 '17 at 5:04
  • $\begingroup$ I think I will go by what Bak and Newman do and assume that f must be bilinear to start off with. If so, it certainly can't have a singularity on the boundary because the image of the open unit disc is bounded by assumption. $\endgroup$ – Acton Dec 11 '17 at 5:19

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