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.


1 Answer 1


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).

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    $\begingroup$ That's right: any space with a continuous multiplication law that makes the set of path-components into a group must be such that left or right multiplication by a point is a homotopy equivalence. $\endgroup$ Aug 8, 2010 at 14:37
  • $\begingroup$ Ah, right. For some reasone I was thinking about the conjugation action instead of simply left and right translation. Thanks. $\endgroup$ Aug 8, 2010 at 14:48
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    $\begingroup$ Tom's comment is addressed by this question on math.SE: math.stackexchange.com/questions/731040/… $\endgroup$ Jun 10, 2014 at 19:09

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