@DanielAsimov Is there a homeomorphism of the disk extendable to a continuous map on the boundary which is not a homeomorphism(However its degree must be perhaps 1 according to the concept of "fundanmental group at infinity" discussed above

@HenrikRüping Thanks for introducing me the "Fundamental group at infinity" I just find a paper with the same title 1996, Topology. Befor I read it I wonder how the rotation number can reflect in fundamental group. I think the degree reflect on fundamental and homology not rotation number?

Correction: Self homeomorphism conjugate to an extendable homeomorphism

@VilleSalo So one may think to a refine version of cobordism theory: Assignment of a group to every cobordant triple $(M,N,W)$, with $\partial W=M \sqcup N$ the normal subgrouo H iof $G=homeo(W^\circ)$ consisting of self homeomorphisms extendable to the boundary

@HenrikRüping in 1 dim. impossible since the rotation number determine the conjugacy class and you said the rotation number is well defined but what about higher dimension?

@VilleSalo "a conjugacy need not extend to the boundary either " Yes that is the point thanks. But I am not aware of two extendable homemorphism which are conjugate on the interior but not on the boundary. it seems that

I am aware of invariance of rotation number under conjugation but the case of non invariance $X\setminus K$ provoke the obsession that: is the limit $\frac{F^n(x)-x}{n}$ really exists? Any reference?

what can be said about the structure of $G/H$? Is there a Lie group (finite or infinite dimensional) structure on this group?

The reason for the obsession: The classical rotation number is sensitive to datas one can not replace a home f with fg

I think for every manifold $M$ with boundary one may associate a normal subgroup $H$ of $G=Home(M^{\circ})$ consist of all homeomorphism conjugate to an extendable self homeomorphism of $M^{\circ}$

@VilleSalo So we take limit of the rotation number as circles approach to infinity. Is not necessary to assume that $X\setminus K$ is $f$ invariant. I guess the answer is no but it is a kind of caution and math obsession

@DanielAsimov Is not even conjugate to a extendable homeomorphism?