# Uniformisation for non simple closed curves

Given a simple closed curve in the plane, there is a homeomorphism from the unit open disk to the interior of the curve. The homeomorphism can be taken conformal, this is the uniformisation theorem.

Is there a version of this result for closed curves that are not simple?

For example, given a $C^1$ closed curve $\gamma: S^1 \rightarrow {\bf R}^2$ with finitely many self-intersections, all of them transverse, is there a continuous map (maybe even conformal) from the unit open disk to the plane, such that the number of preimages of any point in ${\bf R}^2\backslash \gamma(S^1)$ is equal to the absolute value of the number of turns the curve makes around the point?

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• What would such a map look like for the figure-8 curve? – Mikhail Katz Jul 6 '17 at 14:03
• @Katz Twist the disk in three space, project it on a plane. So there is a line dividing the disk that projects exactly on the double point. Perhaps asking for all the number of turns to be of the same sign is a necessary condition for a conformal map. – coudy Jul 6 '17 at 14:11
• @coudy: The map you describe in the comment is not conformal, and not even open. Of course there is always a continuous map: a continuous map of a circle extends continuously in the disk. So what are you really asking? – Alexandre Eremenko Jul 6 '17 at 20:49
• @Eremenko. An arbitrary extension will not necessarily give the expected multiplicity. Read again the question and look at the diagram. – coudy Jul 7 '17 at 6:19
• @goette. The question is the following. Given a C1 closed curve γ:S1→R2 with finitely many self-intersections, all of them transverse, is there a continuous map (maybe even conformal) from the unit open disk to the plane, such that the number of preimages of any point in R2∖γ(S1) is equal to the absolute value of the number of turns the curve makes around the point? – coudy Jul 7 '17 at 19:50