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

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.

• You mean the circle $S^1$, right? Anyway, any connected component of $\Omega S^1$ has $\pi_i = 0$ for all $i > 0$ and is thus contractible (by, for example, Whitehead's theorem). – anomaly Feb 8 '16 at 4:34
• You can show that $\Omega S^1$ has the homotopy type of a CW-complex. Therefore, since any path component is weakly contractible (i.e., has trivial homotopy groups), it is also contractible by Whitehead's theorem. – Nitrogen Feb 8 '16 at 4:40
• Thanks. Yeah in Milnor's : on spaces having the homotopy type of a CW complex, he gives the result(1959). – Hari Rau-Murthy Feb 8 '16 at 4:46
• You have to be careful about basepoints, $\Omega S^1$ is not path-connected so $\pi_i(\Omega S^1)$ (for $i \ge 1$) is either ill-defined or not really what you want. – Najib Idrissi Feb 8 '16 at 8:51

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.