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I have a proof of the following theorem but would like to know whether there is a more elegant or simple proof. Can you prove or disprove it please (showing steps)?

Given a non-constant mereomorphic function $f$ then there exists at least one continuous loop over the extended complex plane $g$ such that $fg$ maps the reals to the reals bijectively.

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I take it the first $g$ is the loop while the second $g$ is a map $\mathbb{R}\to\text{loop}$. Essentially you need to show that the preimage of $\mathbb{R}$ under $f$ is a closed loop in $\mathbb{C}_\infty$. –  anon Aug 15 '11 at 0:38
    
Yes. Im abusing notation as a loop is equipped with a parameterizarion. Also it is sufficient to show the preimage of the reals under $f$ contains aleast one loop (As there may be more than one). –  Harry Barber Aug 15 '11 at 0:49
    
Instead of "at least one continuous loop over the extended complex plane $g$", I'd write "at least one continuous loop $g$ over the extended complex plane". You presumably don't mean that $g$ is the extended complex plane. –  Michael Hardy Aug 15 '11 at 2:08
    
Another question by the OP included some discussion of this problem: math.stackexchange.com/questions/55864/…. –  Jonas Meyer Aug 15 '11 at 2:09
    
Bump. Yes Jonas, I am taking the advice from that thread and seeing whether anyone can prove the theorem relatively easily. :D –  Harry Barber Aug 15 '11 at 16:01
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1 Answer 1

Actually, your claim as stated is wrong even for rational functions.

Indeed, consider the function

$$f:z\mapsto i\cdot\frac{z^2-1}{z^2+1}.$$

The preimage of the real axis (including $\infty$) under this map is the unit circle $\mathbb{T}$.

However, the map $f:\mathbb{T} \to \mathbb{R} \cup \{\infty\}$ is not injective on the unit circle (it is a 2-1 covering map).

For meromorphic functions, you can go even further: Take the map $$f:z\mapsto i\cdot \frac{e^z-1}{e^z+1}.$$ (Can you spot a pattern?) The preimage of the real axis here is just the imaginary axis. So once more, this preimage is a simple closed curve when we add in $\infty$, but the map is an infinite-to-one covering map.

However, we can prove the following. I will replace the extended real axis by the unit circle for convenience (in order to get the original statement, just compose with a Möbius transformation as in the examples above).

Theorem. Let $f$ be a nonconstant meromorphic function. Then there is a nontrivial closed curve $$\gamma\subset f^{-1}(\mathbb{T})\cup\{\infty\}$$ such that $f(\gamma\cap\mathbb{C})$ is a dense subset of $\mathbb{T}$.

Sketch of proof. Let $D$ be the unit disk, and let $V$ be a connected component of $f^{-1}(D)$.

If $V$ is not simply connected, let $\gamma$ be the boundary of one of the complementary components of $V$. It should be easy to see that then $\gamma$ is mapped to $\mathbb{T}$ as a finite-degree covering map.

So suppose that $V$ is simply connected. It is easy to see that the boundary of $V$ is locally connected near every finite point. Since a continuum cannot fail to be locally connected at only one point, it follows that the boundary of $V$ is locally connected. By Carathéodory's theorem, the boundary is the image of a continuous curve $\gamma:\mathbb{T}\to \partial V$. To see that $f(\gamma\cap\mathbb{C})$ is dense in the unit circle, we can simply apply the Gross star theorem. This theorem says that a branch of the inverse of a meromorphic function can be analytically continued along almost every radial ray. This completes the proof.

One can do a closer analysis of the mapping behavior of $f$ on the curve $\gamma$. Of course if $f$ is rational, then we can ensure that $f$ maps $\gamma$ to $\mathbb{T}$ as a finite-degree covering map.

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