# self-study problem: the behavior of conformal mapping at the Boundary

I am reading Rudin's book, Real and Complex Analysis, and face a problem about the proof of Theorem 14.18. Here is my problem:

The hypothesis of the theorem is "$f$ is a conformal mapping of $\Omega$ onto $U$". However, in the proof, it seems that the author blindly assumed $f$ is one-to one, because he said "Let $g$ be the inverse of $f$".

I don't know why $f$ has a inverse, so I check the definition of conformal mappings. In the book, conformal mappings are defined as holomorphic functions with nonvanishing derivative. I know a one-to-one holomorphic function must have a nonvanishing derivative. However, nonvanishing derivative generally cannot imply a holomorphic function is one-to-one, but imply it is locally one-to-one.

Maybe it's a very stupid problem. Any explanations will be appreciated.

BTW: I've checked lots of textbooks relevant to the theorem. However, some books define confomal mappings as one-to one holomorphic functions with nonvanishing derivative. Others state $f$ is biholomorphic in the hypothesis of the theorem. Nothing is helpful for me to understand the theorem in Rudin's book.

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Considering "one-to-one" to be a part of the definition of "conformal mapping from one region to another", even if not part of the definition of "conformal mapping", seems to be a cultural convention. Or could that be an impression I somehow formed mistakenly? – Michael Hardy Dec 11 '11 at 17:28
@MichaelHardy: I think you raise a good point. But as Y. Fan says, that convention conflicts with the actual definition that Rudin gives in the book, and he never mentions a switch to adopting the convention. Therefore it is in my opinion an actual oversight, albeit an understandable and easily fixable one. – Jonas Meyer Dec 11 '11 at 20:45