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I am given to understand that the group of diffeomorphisms of the unit circle, $operatorname{Diff}(mathbb{S}^1)$, has two connected components, $operatorname{Diff}^+(mathbb{S}^1)$ and $operatorname{Diff}^-(mathbb{S}^1)$, the diffeomorphisms that preserve or reverse the canonical (counterclockwise) orientation, respectively.

Question 1: How does one prove that?

Question 2: Given $Phi, Psi in operatorname{Diff}^+(mathbb{S^1})$, can one construct an explicit path joining them?

I'd be satisfied already with a proof that we can join $Psi, Phi in operatorname{Diff}^+ (mathbb{S}^1)$, without giving the path explicitly. I tried the obvious path, $t mapsto dfrac{tPhi + (1-t)Psi}{|tPhi + (1-t)Psi|}$, but that doesn't seem to work.

Thanks.

Question 2:

As Mariano points out, we can lift $Phi in operatorname{Diff}^+{(mathbb S^1)}$ uniquely to a diffeomorphism $phi : mathbb{R} to mathbb{R}$ such that $phi(0) in [0,1)$ and $phi(x+1) = phi(x) + 1$ for all $x in mathbb{R}$. Lift $Psi$ similarly to $psi$. For $t in [0,1]$ the map $gamma_t = (1-t) phi + tpsi$ is a diffeomorphism $mathbb{R} to mathbb{R}$ (it is strictly monotonically increasing because $ gamma_{t}^prime(x) gt 0$ for all $x in mathbb{R}$ and all $t in [0,1]$) such that $gamma_t(0) in [0,1)$ and $gamma_t(x+1) = gamma_t(x)+1$. Thus $gamma_t$ descends to a diffeomorphism $Gamma_t in operatorname{Diff}^+(mathbb{S}^1)$, $Gamma_0 = Phi$ and $Gamma_1=Psi$. It is straightforwad to check that $t mapsto Gamma_t$ is a continuous path.

What we exploited here is that we have a short exact sequence (in fact a central extension)
$$
0 to mathbb{Z} to operatorname{Diff}_{mathbb{Z}}{(mathbb{R})} to operatorname{Diff}^+{(mathbb{S}^1)} to 1
$$
where $operatorname{Diff}_{mathbb{Z}}{(mathbb{R})}$ denotes the group of diffeomorphisms $mathbb{R} to mathbb{R}$ commuting with the shift $tau(x) = x+1$ and $mathbb{Z} = langle tau rangle$ and the lift $Phi mapsto phi$ is a section of this central extension.

Question 1:

It is clear that a diffeomorphism $mathbb{S}^1 to mathbb{S}^1$ either preserves or reverses orientation and that the orientation-preserving diffeomorphisms $operatorname{Diff}^+{(mathbb{S}^1)}$ form a normal subgroup of $operatorname{Diff}{(mathbb{S}^1)}$. Now simply use the conjugation diffeomorphism $z mapsto bar{z}$ to see that $operatorname{Diff}^+{(mathbb{S}^1)}$ has index 2.

For an excellent introduction to $operatorname{Homeo}^+{(mathbb{S}^1)}$ and $operatorname{Diff}^+{(mathbb{S}^1)}$ and their subgroups I recommend Étienne Ghys's article
Groups acting on the circle, Enseign. Math. (2) 47 (2001), no. 3–4, 329–407, MR2111644. The short exact sequence mentioned above plays a central rôle in the theory.

For more on the diffeomorphism group of the circle, I recommend consulting the work of Andrés Navas, and, of course, the classic article:

Michael R. Herman, Sur la conjugaison différentiable des difféomorphismes du cercle à des rotations, Publications Mathématiques de l'IHÉS, 49 (1979), p. 5–233.