# Any orientation preserving self-homeomorphism of $S^2$ is isotopic to identity

I want to show that any orientation preserving self-homeomorphism of a 2-sphere $S^2$ is isotopic to identity.

Any help or reference is appreciated.

Edit; I want to show this to prove the following. Suppose we have two solid torus and we have a homeomorphism of the boundaries. The manifold obtained by indentifying boundaries via the homeomorphism depends only on the image of the meridian. To show this, first cut out the cylinder neighborhood of a meridian and glue it to the other solid torus. The reminder is homeomorphic to $B^3$. So if I can prove the question above, I can finish this proof.

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Why do you want to show that? What have you tried? – Henning Makholm Apr 16 '12 at 21:13
@Henning I edited to include the motivation of the question. I don't know exactlly where to start. – Primo Apr 16 '12 at 21:28
The remainder is a 3-ball, not a 2-sphere!! – user641 Apr 17 '12 at 5:15
And it is very easy to see there is only one way to glue a 3-ball... – user641 Apr 17 '12 at 5:15
Again, I will say this: since the remainder is a 3-ball, you don't need any of this: it is very easy to see you can glue the 3-ball in in a unique way. Nothing about the 2-sphere is relevant to your main question. – user641 Apr 17 '12 at 7:11

In particular, this paper proves that the space of orientation-preserving diffeomorphisms of $S^2$ that fix three points on the circle is contractible. Note that a path in the space of diffeomorphisms is precisely an isotopy.