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.



Are the path components of a loop space homotopy equivalent?
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).


