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Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a meromorphic function such that $f'(z) \ne 0$ for all $z \in \mathbb{C}$. Let $\Gamma \subset \mathbb{C}$ be a curve which has no self intersections. If we assume that for any $\omega \in \Gamma$, $f^{-1}(\{\omega\}) \ne \emptyset$, then my question is that, can we find a curve $\gamma \subset \mathbb{C}$ such that $f$ is one-to-one on $\gamma$ and that $f(\gamma)=\Gamma$?

I believe this is true because such function $f$ is locally one-to-one, and even if points on $\Gamma$ can be possibly taken infinitely many times on the complex plane $\mathbb{C}$, we can always choose a certain "branch" such that the restricition of $f$ on such a "branch" is one-to-one. But I'm not sure how to give a rigorous argument to validate the existence of such "branch". Any comments or suggestions will be fully appreciated.

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No, this is not true. The simplest example is $f(z)=\int_0^ze^{-\zeta^2}d\zeta$. Preimage of the real line consists of infinitely many curves, each of them is mapped homeomorphically onto one or two intervals of the three intervals $(-\infty,-\pi/2),\; (-\pi/2,\pi/2),\; (\pi/2,+\infty)$. But none of the curves is mapped on the whole real line. The picture of these curves is given in Fig. 24, section 253, Chapter XI of the book Nevanlinna, Analytic functions. The book does not contain complete proofs; they are given in Nenvanlinna's paper Über Riemannsche Flächen mit endlich vielen Windungspunkten, Acta Math. 58 (1932), no. 1, 295–373.

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  • $\begingroup$ Thank you very much! This answer is definitely elegant but certainly beyond my knowledge... Could you elaborate more why $f(z)$ has all of these properties, or just provide a reference for me to check those out? $\endgroup$ – student May 17 at 3:03
  • $\begingroup$ @student: I added some references. $\endgroup$ – Alexandre Eremenko May 17 at 4:27
  • $\begingroup$ Thank you very much! $\endgroup$ – student May 17 at 16:16

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