If $X$ is a based space, then we have $\pi_1(X) \cong \pi_0(\Omega X)$. This is to say we can identify elements in the fundamental group of $X$ with path components of the first loop space of $X$. My question is this: do all the path components of $\Omega X$ necessarily have the same homotopy type?
I ask because when you iterate this idea by considering $\pi_2(X) \cong \pi_1(\Omega X) \cong \pi_0(\Omega^2 X)$, the space $\Omega^2 X$ only sees the component of $\Omega X$ containing the basepoint (in this case the constant map). So when I imagine climbing up the tower of $X$'s homotopy groups, at each stage we apply the loop functor and discard all the information in the non-basepoint components.
If all the components of $\Omega X$ are homotopy equivalent, then this is clearly no loss of information, since we can identify the homotopy groups of $X$ with the (properly shifted) homotopy groups of any of the connected components. Otherwise, don't we generate much more information about $X$ by looking at the homotopy types of all the path components of $\Omega X$, $\Omega^2 X$, etc.?
I keep thinking there might be some argument about a certain map being a fibration, and this giving an equivalence between the various path components of $\Omega X$, realized as fibers of the map, or something like this, but I haven't found it yet, and I can think of no canonical maps which might serve as the desired equivalences.
Best Answer
Does the following work?
Let $A$ and $B$ be components of $\Omega X$ and assume that $A$ is the component containing the trivial path based at $x_0$. Let $f$ be any element of $B$. Then $f$ is a path from $x_0$ to $x_0$. There's a map $\phi:A\to B$ obtained by following a path in $A$ by $f$. Similarly there's $\psi:B\to A$ obtained by following a path by the reverse of $f$.
Surely $\phi$ and $\psi$ are homotopy inverses of each other?
(Warning: I'm not a topologist by trade, so I fear that I may have missed something subtle, or indeed something obvious, here).