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  • $\begingroup$ How do you know the arc has ends? $\endgroup$ Commented Mar 8, 2011 at 4:14
  • $\begingroup$ Based on the stackexchange question, I also assumed the arc is a compact one. $\endgroup$ Commented Mar 8, 2011 at 4:19
  • $\begingroup$ @Richard: in case your response is aimed at me, what I mean is in Stillwell's argument "our arc on a strip in the plane" is an arc in the universal cover of a twice punctured sphere -- this cover is homeomorphic to $\mathbb R^2$, so showing the arc has endpoints on the boundary is a non-trivial problem (it's easy to construct ones that don't, even if they come from embedded intervals in $S^2$ via this construction). $\endgroup$ Commented Mar 8, 2011 at 4:24
  • $\begingroup$ Oh, I think I see how one can tweak Stillwell's argument to make it work -- the lifted arc is in $(-1,1) \times \mathbb R$ which you could compactify to $[-1,1] \times [-\infty,\infty]$. Crush $\{-1\}\times [-\infty,\infty]$ and $\{1\}\times [-\infty,\infty]$ to points and identify the quotient space with a compact disc -- the arc has ends in this quotient space. $\endgroup$ Commented Mar 8, 2011 at 4:33
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    $\begingroup$ Actually, if the arc goes from $0$ to $\infty$ in $\hat{\mathbb C}$, then just take the preimage under $z \mapsto z^2$: it's a circle. It's nice, but it reduces the problem to another problem that ab initio is just as hard, just better known. $\endgroup$ Commented Mar 8, 2011 at 4:46