Are the path connected components of $\Omega S_1$ contractible?

Question

Let $\Omega S_1$ be the space of loops of $S_1$ based at $x_0 \in S_1$ with the topology of uniform convergence. We know that the path connected components of this space are in one to one correspondence with $\pi_1(S_1)$. Concretely this is because the a path in $\Omega S_1$ between two maps $\phi_a: S_1 \to S_1$ and $\phi_b: S_1 \to S_1$ is a homotopy rel $x_0$ between these two maps.

Now I would like to know if the path connected components of $\Omega S_1$ are contractible. As a first indication, any loop that is contained in a path connected component is contractible in the loopspace: any loop must be contained in a path connected component and $\pi_1(\Omega^1 S_1)=\pi_2(S_1)=0$ where the last equality is true because any map from $S_2 \to S_1$ has a lift to $S_1$'s universal cover.

More generally we have that $\pi_i(\Omega^1 S_1)=\pi_{i+1} (S_1)=0$ (for $i>0$)

So I am pretty sure the answer is yes. But I am not sure how to prove it.

Answer 1

For a homotopist's proof see the comment I made above. There is a simpler proof however.

Take a component $X$ of $\Omega S^1$. It corresponds to some integer $n$ because $\pi_0(\Omega S^1)\cong \Bbb{Z}$. That means that $X$ is the subspace of loops going $n$ times around the circle.

By the general theory of universal covers, you can prove that $X$ is homeomorphic to the space $Y$ of paths in $\Bbb{R}$ going from $0$ to $n$.

So you just have to prove that $Y$ is contractible, which can easily be seen using the fact that $\Bbb{R}$ is convex.