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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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    If my understand$i$$n$g (above) is correct, then I am a bit dubious about the statement. If the inverse of the meromorphic function has no indirect singularities (in the sense of classical function theory), then the claim should be easy. However, for maps with indirect singularities, I see _a priori_ no reason why the statement should be true.2011-10-21

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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.