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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 $\mathbb{C}_\infty$.2011-08-15
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    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).2011-08-15
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    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.2011-08-15
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    Another question by the OP included some discussion of this problem: http://math.stackexchange.com/questions/55864/how-does-one-know-that-a-theorem-is-strong-enough-to-publish.2011-08-15
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    Bump. Yes Jonas, I am taking the advice from that thread and seeing whether anyone can prove the theorem relatively easily. :D2011-08-15
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    There is some possible confusion about the statement, since $f$ is not defined at $\infty$. Could you clearify if I understand you correctly? If I do, your claim is as follows: For every meromorphic function $f$, there exists a continuous map $\alpha$ from the extended real axis to the extended complex plane such that the composition $f\alpha$ is defined and equals the identity on a dense subset of the reals.2011-10-21
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    If my understanding (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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