Is it true that the loop space of a circle is contractible? Consider the long exact sequence in homotopy for the path fibration $\Omega S^1 \rightarrow \ast \rightarrow S^1$ shows all homotopy groups of the loop space to be zero, and then Whitehead's theorem kicks in and tells us that $\Omega S^1$ is contractible.
I can't see any flaw in this argument, but it's really rankling against my intuition of what a loop space "is". For instance, we know that $S^1$ has $\mathbb{Z}$ many loops up to homotopy, so surely it should have a fairly complicated loop space? So I suppose that there are two possible approaches to answering this question: either point out where I'm going wrong in my mathematics, or help rectify my intuition.
Secondly, does this generalise to the $n$-fold loop space of an $n$-sphere? I think it does; I've used the Serre spectral sequence to compute the cohomology rings of loop spaces of spheres and they all die at $\Omega^n S^n$.
Sorry for the "please check my working" nature of the question; it's sufficiently early in my mathematical life that I'm never convinced that there's no errors in my working (especially not with spectral sequences, which still feel a bit like doing magic to me). It was a lot easier to be (mathematically) confident as an undergraduate, when you could check your working against solutions sheets…
Best Answer
$\Omega S^1\cong \mathbb Z$ — your argument is almost correct: each connected component is contractible, but there are countably many such components ($\pi_0(\Omega S^1)=\pi_1(S^1)=\mathbb Z$).
As for $\Omega^n S^n$, these spaces have very non-trivial homotopy type: $\pi_k(\Omega^n S^n)=\pi_{n+k}S^n$ and there are a lot of non-trivial homotopy groups of spheres (say, $\pi_1(\Omega^2 S^2)=\pi_3(S^2)=\mathbb Z$, $\pi_1(\Omega^3 S^3)=\pi_4(S^3)=\mathbb Z/2\mathbb Z$ etc).
As for the Serre spectral sequence, maybe you made the standard mistake: when computing, say, $H(\Omega S^3)$ one sees that there is an element $x$ s.t. $d_2(x)=[S^3]$ and concludes that (since powers of $x$ "obviously" kill all cohomology classes from $E_2$) $H(\Omega S^3)$ "is" $\mathbb Z[x]$. But actually $d_2(x^n)=\mathbf{n}x[S^3]$, so powers of $x$ don't kill all classes: there also should exist classes $x^n/n!$ and the answer is not $\mathbb Z[X]$ but $\Gamma[x]$.