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Consider the closed two dimensional disk and two fixed points $A$ and $B$ on the boundary. I am looking for a reference for the following fact: a simple topological curve with end points $A$ and $B$ is unique up to an ambient isotopy, relative to the boundary of the disk.

There is a result that says that two simple closed curves in a compact manifold are isotopic if and only if they are homotopic. Does this apply to compact manifolds with boundary too, like the closed disk? Is the isotopy an ambient isotopy? If so, then I guess one can use that to prove the above statement in the above paragraph.

Thanks!

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    $\begingroup$ For theorems of this sort, look at Farb and Margalit's Primer on Mapping Class Groups. There are also exercises to guide you through the proofs in Rolfsen. In general, (for arbitrary dimension) the isotopy can't be an ambient isotopy as that would imply that all knots are equivalent to the unknot. $\endgroup$ Jul 3, 2014 at 14:52
  • $\begingroup$ Thanks. I should have mentioned that the disk is two dimensional. I guess in that case the result is true? I tried looking at the Primer on Mapping Class Groups, which is where I got the result on homotopic curves in a surface being isotopic. $\endgroup$
    – guest84
    Jul 3, 2014 at 15:01
  • $\begingroup$ The last paragraph on p.35 of the Primer explicitly states that "homotopy implies isotopy" also works for simple arcs between boundary points. You would have to tweak the proof of Proposition 1.10 just a little. $\endgroup$ Jul 3, 2014 at 15:20
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    $\begingroup$ The result itself is in the paper by David Epstein "curves on 2-manifolds and isotopes" from 1966 (it is an application of Schoenflies theorem). $\endgroup$
    – Misha
    Jul 3, 2014 at 15:28
  • $\begingroup$ Indeed, it is mentioned in the Primer! Thank you, I had not noticed it. Is that an isotopy relative to the boundary, though? In other words, will the end points of the arc remain fixed throughout the isotopy? I guess, if the arcs are smooth, then they are also ambient isotopic, right? $\endgroup$
    – guest84
    Jul 3, 2014 at 15:32

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