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Suppose you have $\gamma(t):[0,1]\rightarrow \mathbb{C}$ simple piecewise smooth, $\gamma(0) = 0$ and $\gamma(1)=1$. Does there exist $\eta:[0,1]\rightarrow \mathbb{C}$, another simple piecewise smooth curve, with $\eta(0) = 0$ and $\eta(1)$ a point on the unit circle, such that $\gamma$ and $\eta$ only intersect at the origin?

My attempt: I tried approaching this constructively, but I couldn't seem to get my hands on anything useful. The best alternative I could come up with was trying to use the Riemann mapping theorem to find a conformal map $f: \mathbb{D}\setminus\gamma \rightarrow \mathbb{D}$. If $f$ extends continuously to the boundary, then I could use a chord $A$ which joins $f(0)$ to $f(-1)$, and then let $\eta = f^{-1}(A)$. But, even then, I'm not sure if I can guarantee that the resulting $\eta$ has nonzero one-sided derivatives at the endpoints. So my strategy only works in nice situations (as far as I know).

Is there a general topological approach to this? Can my Riemann mapping strategy be made to work, or is it hopeless?

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  • $\begingroup$ (Nitpick: $\gamma$ has to be injective.) $\endgroup$ – Chris Culter Aug 24 '15 at 7:58
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    $\begingroup$ Judging by your attempt, you want $\eta(1)$ to be a point on the unit circle, not "on the unit disk". Is that right? $\endgroup$ – Daniel Fischer Aug 24 '15 at 8:10
  • $\begingroup$ Ah, right. Sorry! Also corrected. $\endgroup$ – Josh Keneda Aug 24 '15 at 8:31
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    $\begingroup$ This is just an idea: It seems intuitively that the path's benign properties should guarantee that you can define a path at distance $\epsilon$ "to the left" of $\gamma$, and that the infimum of $\epsilon$ such that this path intersects $\gamma$ should be non-zero. If so, it seems it should be possible to hook such a path up to the origin and the unit circle for sufficiently small $\epsilon$? $\endgroup$ – joriki Aug 24 '15 at 8:32
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    $\begingroup$ @joriki Ah. I haven't met those before, so I'll read up and see if I can complete your argument. Thanks for your help! $\endgroup$ – Josh Keneda Aug 24 '15 at 12:18

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