# Space of homeomorphisms Homeo$(S^1)$ of $S^1$ deformation retracts onto $O(2)$

How can we prove that the space of homeomorphisms Homeo$(S^1)$ of $S^1$ (strong) deformation retracts onto the orthogonal group $O(2)$?

I know that this result is proved by Hellmuth Kneser in his paper Die Deformationssätze der einfach zusammenhängenden Flächen.

I want to learn somewhat more elementary proof if possible, or at least the idea of the proof.

Added Later (The topology on the space of homeomorphisms Homeo$(S^1)$ of $S^1$). This is footnote 7 from Hutchings' lecture notes, Introduction to homotopy groups and obstruction theory:

We topologize the space Maps$(X, Y)$ of continuous maps from $X$ to $Y$ using the compact-open topology. For details on this see e.g. Hatcher. A key property of this topology is that if $Y$ is locally compact, then a map $X \rightarrow$ Maps$(Y,Z)$ is continuous iff the corresponding map $X \times Y \rightarrow Z$ is continuous.

• What is the topology on Homeo($S^1$)? Commented Mar 4, 2013 at 22:31
• @StefanH. I edited the post to answer your question. Thanks! Commented Mar 4, 2013 at 23:00
• So it is just a metric space with the sup-Norm. Commented Mar 4, 2013 at 23:14
• What do you mean by "somewhat more elementary"? On page 367 Kneser writes down an explicit deformation retraction: $H(x,t) = (1-t)f(x) \pm tx + tf(0)$ where the sign depends on whether $f$ preserves or reverses orientation. The idea is to view a homeomorphism of $S^1$ as a strictly monotonic function $f \colon \mathbb{R} \to \mathbb{R}$ subject to the condition $f(x+1) = f(x) \pm 1$. Commented Mar 5, 2013 at 5:34

The following is essentially a fleshed-out version of the proof of Proposition 4.2 in Ghys's beautiful article Groups acting on the circle.

It is convenient to translate the question into a question about continuous functions on $$\mathbb{R}$$. Viewing the circle $$S^1 = \mathbb{R}/\mathbb{Z}$$ as a quotient of $$\mathbb{R}$$, a continuous function $$f\colon S^1 \to S^1$$ can be lifted over the covering projection $$\pi \colon \mathbb{R} \to S^1$$ to a continuous function $$F \colon \mathbb{R} \to \mathbb{R}$$ such that $$f \circ \pi = \pi \circ F$$: on the fundamental domain $$[0,1)$$ the function $$F$$ is determined by $$f$$ up to addition of an integer, and once that integer is chosen, there is only one way to extend $$F$$ continuously to all of $$\mathbb{R}$$.

A continuous function $$F \colon \mathbb{R} \to \mathbb{R}$$ is the lift of a continuous function $$f \colon S^1 \to S^1$$ if and only if there is $$d \in \mathbb{Z}$$ such that for all $$x$$ the equation $$F(x+1) = F(x) + d$$ holds. It is not hard to see that $$d = \deg{f}$$. Using this [or by elementary considerations with the intermediate value theorem] we see that $$f$$ can only be a homeomorphism if $$d = \pm 1$$: the lifts of a homeomorphism must satisfy $$F(x+1) = F(x) \pm 1$$. The lift $$F$$ of a homeomorphism $$f$$ is a homeomorphism because continuous bijections $$\mathbb{R} \to \mathbb{R}$$ have continuous inverses.

Let $$\tilde{H} = \operatorname{Homeo}_\mathbb{Z}(\mathbb{R})$$ be the group of homeomorphisms $$F \colon \mathbb{R} \to \mathbb{R}$$ such that $$F(x+1) = F(x) +1$$. The compact-open topology on $$\tilde{H}$$ coincides with the uniform topology and is metrized by $$d(F,G) = \sup_{x \in [0,1]} \lvert F(x) - G(x)\rvert$$ since $$F-G$$ is $$1$$-periodic. If $$F,G \in \tilde{H}$$ then their convex combinations $$(1-t)F + tG$$ belong to $$\tilde{H}$$ for all $$t \in [0,1]$$: they are strictly increasing and surjective. The map $$h\colon [0,1] \times \tilde{H} \times \tilde{H} \longrightarrow \tilde{H}, \quad(t,F,G) \longmapsto (1-t)F + tG$$ is clearly continuous. This shows that $$\tilde{H}$$ is contractible, as we can take $$G$$ to be the identity $$G(x) = x$$ and for all $$F \in \tilde{H}$$ we have $$h(0,F,G) = F$$ and $$h(1,F,G) = G$$.

The map $$F \mapsto f$$ sending $$F \in \tilde{H}$$ to the homeomorphism of $$S^1$$ it covers is a continuous homomorphism $$p\colon \tilde{H} \to H = \operatorname{Homeo}_+(S^1)$$ onto the group of orientation-preserving homeomorphisms (i.e., those $$f \colon S^1 \to S^1$$ preserving the cyclic order of triples of pairwise distinct points on the circle). The kernel of $$p$$ can be identified with $$\mathbb{Z} = \langle \tau \rangle$$ generated by the unit translation $$\tau(x) = x + 1$$. In fact, $$p$$ identifies $$\tilde{H}$$ with the universal covering group of $$H$$.

Observe that $$f = p(F)$$ is a rotation with angle $$[\alpha] \in \mathbb{R/Z}$$ if and only if $$F(x) = x + \alpha$$ is a translation.

Now notice that each $$F \in \tilde{H}$$ can be uniquely written as $$F(x) = x + \alpha_F + \varphi_F(x)$$ with $$\alpha_F = \int_{0}^1 [F(x)-x]\,dx \in \mathbb{R}$$ and $$\varphi(x) = F(x) - x -\alpha$$ is a $$1$$-periodic function with mean zero: $$\int_{0}^1 \varphi(t)\,dt = 0$$. Since $$F_n \to F$$ in $$\tilde{H}$$ means uniform convergence, we see that $$\alpha_F$$ and $$\varphi_F$$ depend continuously on $$F$$. Thus, the map $$\tilde{k} \colon [0,1] \times \tilde{H} \longrightarrow \tilde{H} \quad \tilde{k}(t,F) = x + \alpha_F + (1-t)\varphi_F$$ is continuous and $$\tilde{k}(0,F) = F$$ while $$\tilde{k}(1,F) = x+\alpha_F$$ covers a rotation, i.e., an element of $$SO(2)$$. If $$F(x) = x + \alpha$$ is a translation then $$\tilde k(t,F) = F$$ for all $$t \in [0,1]$$, so $$\tilde k$$ is constant on the group of translations.

For two lifts $$F_1(x) = x + \alpha_{F_1} + \varphi_{F_1}(x)$$ and $$F_{2}(x) = x + \alpha_{F_2} + \varphi_{F_2}(x)$$ of the homeomorphism $$f \colon S^1 \to S^1$$ we have $$\varphi_{F_1} = \varphi_{F_2}$$ and $$\alpha_{F_1} - \alpha_{F_2} \in \mathbb{Z}$$. This implies that $$\tilde{k}(t,F_1)$$ and $$\tilde{k}(t,F_2)$$ cover the same homeomorphism of $$S^1$$ and hence $$\tilde{k}$$ yields a continuous map $$k\colon [0,1] \times H \to H$$ which is a deformation retraction from $$\operatorname{Homeo}_+(S^1)$$ to $$SO(2)$$.

The argument for the group of all homeomorphisms of $$S^1$$ is essentially the same: If $$f$$ reverses orientation then its lifts are of the form $$F(x+1) = F(x) - 1$$ and $$F$$ is strictly decreasing. Now define $$\alpha_F = \int_{0}^1 [F(x) + x]\,dx$$ and $$\varphi_F = F + x - \alpha_F$$ to obtain a homotopy from $$F$$ to $$\alpha_F - x$$, covering the reflection at the line with angle $$\alpha_F/2$$.

• Great answer, thank you! Commented Mar 8, 2013 at 18:32
• The link is not working. Commented Jul 29, 2018 at 10:52