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Let $\mathbb{D}=\{z\in \mathbb{R}^2\mid |z|<1\}$

Is it true to say that every homeomorphism of $\mathbb{D}$ is conjugate to a self homeomrphism of the disk extendable to a homeomorphism of $\bar{\mathbb{D}}$?

Is there a self homeomorphism of the disk which is conjugate to two self homeomorphism $f$ and $g$ of the disk extendable to homeomorphisms of closed disk with different rotation number?

The rotation number is meant for restriction to the boundary

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    $\begingroup$ No. One can find a smooth homeomorphism of the open disk that takes each circle about the origin, say of radius r β‰₯ 1/2, to itself by a rotation by angle πœƒ(r), satisfying πœƒ(r) β†’ ∞ as r β†’ ∞. This cannot be extended to the closed disk. $\endgroup$
    – Daniel Asimov
    Commented Sep 12 at 22:33
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    $\begingroup$ @DanielAsimov Is not even conjugate to a extendable homeomorphism? $\endgroup$
    – Ali Taghavi
    Commented Sep 12 at 22:35
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    $\begingroup$ The "subtlety" as I see it is that a conjugacy need not extend to the boundary either $\endgroup$
    – Ville Salo
    Commented Sep 13 at 5:49
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    $\begingroup$ The issue with this question is really that there are two questions. I believe the first one is subtle, and I dont know how to decide whether @DanielAsimov's example is conjugate to a homeomorphism that extends to the boundary. I have no idea whether this is research level. $\endgroup$
    – HenrikRüping
    Commented Sep 13 at 8:49
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    $\begingroup$ And I can tell you that even the fundamental group is not taught in all first topology courses. Source: I teach the only topology course in our university. (FWIW I did not design this course) $\endgroup$
    – Ville Salo
    Commented Sep 13 at 9:06

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