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I have proved the following Diff$^+(S^1$) is path connected. Now I want to prove it is deformation retracts to $O(2)$.

What I tried is the following:

I define an onto homomorphism $$ f:\text{Diff}^{+}(\mathbb{D}^2)\to \text{Diff}^{+}(S^1) $$ by $$ \phi\mapsto \phi|_{\partial \mathbb{D}^2=S^1}$$ This is onto because any diffeomorphism of $S^1$ can be extended to a diffeomorphism of $\mathbb{D^2}$. The kernel of this homomorphism is $\text{Diff}^+(\mathbb{D}^2\text{rel } \partial)$. After that I don't know what to do.

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marked as duplicate by Andres Mejia, José Carlos Santos, Namaste, Adrian Keister, Alexander Gruber Jul 29 '18 at 5:50

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  • $\begingroup$ If you are using $Diff^+(S^1$, then I think you want $SO(2)$, since $O(2)$ is not path connected and does not act in an orientation-preserving manner. $\endgroup$ – Tyrone Jul 28 '18 at 16:48
  • $\begingroup$ Yeah, that's fine but I want to prove Diff$(S^1)$ is deformation retracts to $O(2)$. $\endgroup$ – Sachchidanand Prasad Jul 28 '18 at 17:01
  • $\begingroup$ Think of this as fibration. If you assume continuous category is same as differential category in low dimension (2&3)...then try to prove that each of the fiber Diff+(Drel\partial) are contractible by using Alexander trick. And thus complete the proof. $\endgroup$ – Anubhav Mukherjee Jul 28 '18 at 17:01
  • $\begingroup$ @SachchidanandPrasad, you may have to rethink, then, since $Diff^+(S^1)$ does not deformation retract onto $O(2)$, but rather onto $SO(2)$. $\endgroup$ – Tyrone Jul 28 '18 at 17:33
  • $\begingroup$ @Tyrone I know that Diff$^+(S^1)$ is deformation reatract to $SO(2)$. But I want to prove the result for Diff$(S^1)$ $\endgroup$ – Sachchidanand Prasad Jul 28 '18 at 19:48
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We will show the result for homotopy equivalences of $S^1$, denoted $Self(S^1)$, and then show that we can restrict down to either the homeomorphism or diffeomorphism group. This is all discussed in Lurie's notes, but I'll reproduce the argument for completeness.

both $Diff(S^1)$ and $Self(S^1)$ can be decomposed into two pieces, orientation preserving and orientation reversing, and likewise so can $O(2)$.

Hence, it is enough to show that $SO(2)$ is a deformation retract of $Diff^+(S^1)$, as noted in the comments.

First, note that there are decompositions $Diff^+(S^1)=Diff_0^+(S^1)SO(2)$ and $Self^+=Self_0^+(S^1)SO(2)$, where the subscript $0$ indicates that they fix a base point of $S^1$. In other words, every diffeomorphism can be obtained from one that fixes a base point, and then rotates appropriately.

First, we can regard $S^1= \mathbb R/\mathbf Z$, so a base-point perserving map $f:S^1 \to S^1$ can be regarded as a map $\tilde{f}:\mathbb R \to \mathbb R$ where $f(x+1)=f(x) \pm d$ and $\tilde{f}(0)=0$ (take the perserved fixed point to be the image of $0$.) Note that $f$ is a homotopy equivalence if and only if $d= \pm 1$, since $d$ is the degree of the map, and since it is orientation perserving, $d=1$.

Hence, we have an identification $Self_0^+(S^1)=\{\tilde{f}:\mathbb R \to \mathbb R \mid \tilde{f}(0)=0, f(x+1)= x+1\}$.

However, this lets us form the straight line homotopy $F_t:=(1-t)\tilde{f}(x)+tx$, which shows the space to be contractible (retractible) (as all functions are homotopic to identity)

$Diff_0^+(S^1) \subset Self_0^+(S^1)$ in the above identification by taking the subset of smooth maps without vanishing derivative. We can follow the same homotopy.

In particular, this shows a retract via the straight line homotopy from $Diff^+(S^1)$ to $SO(2)$

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  • $\begingroup$ Will you please tell me why $\text{Diff}^+(S^1)=\text{Diff}_0^+(S^1)SO(2)$ what is the meaning of $\text{Diff}_0^+(S^1)SO(2)$? $\endgroup$ – Sachchidanand Prasad Jul 29 '18 at 8:04

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