Suppose X is a loop space; by this we mean there is some space $Y$ with $\Omega Y \simeq X$.

Under what assumptions is (the homotopy type of) $Y$ unique?

As has been pointed out below, the homotopy type of $Y$ being determined uniquely is far from true in general. But for connected $Y$, are there conditions we can impose that make it so?

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    $\begingroup$ The question is, probably: «if $\Omega X_1$ and $\Omega X_2$ are homotopy equivalent, are $X_1$ and $X_2$ homotopy equivalent?» $\endgroup$ – Mariano Suárez-Álvarez May 16 '11 at 3:12
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    $\begingroup$ Don't have time to leave a full answer, but if the homotopy groups of X are concentrated in a narrow "stable" range (roughly the top nonzero group below about twice the dimension of the bottom nonzero group) then there is a unique homotopy type. Similarly if Y has very small dimension relative to its connectivity. One can compare the obstruction theory for $Map(Y_1, Y_2)$ with that for $Map(\Omega Y_1, \Omega Y_2)$, together with a comparison of the cohomology of $Y_1$ and $\Omega Y_1$, to get a fuller statement. $\endgroup$ – Tyler Lawson May 16 '11 at 3:57
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    $\begingroup$ Yes, of course you would want $Y$ to be connected to avoid trivialities. $\endgroup$ – Dr Shello May 16 '11 at 6:19
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    $\begingroup$ @Mariano: yes, right. The question is what assumptions are needed to make that true. $\endgroup$ – Dr Shello May 16 '11 at 6:25
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    $\begingroup$ You might also ask a slightly different question: "Given a homotopy equivalence of based loop spaces, how do I tell if it is a loop map?" $\endgroup$ – S. Carnahan May 16 '11 at 8:31

As Ryan points out, if Y is allowed to be disconnected, then there is no hope, since the loop-space construction sees only the connected component of the basepoint. But even if Y is assumed to be connected, it is not unique. For instance, let G and H be two discrete groups whose underlying sets are bijective, but which are not isomorphic. Then as (discrete) topological spaces, we have $G\simeq H$, and so both $K(G,1)$ and $K(H,1)$ are spaces Y such that $\Omega Y \simeq G \simeq H$. But $K(G,1)$ and $K(H,1)$ are not homotopy equivalent unless $G\cong H$ as groups.

What is true, however, is that if we remember the "up-to-coherent-homotopy" multiplication (i.e. "$A_\infty$-structure") on a loop space $\Omega Y$, then the connected space Y is characterized up to homotopy equivalence by $\Omega Y$ and this additional data. For there is a delooping functor "B" from $A_\infty$-spaces to connected spaces, which preserves homotopy equivalence, and such that $B\Omega Y \simeq Y$.

  • $\begingroup$ So this means in general for $Y$ connected, its homotopy type is not unique. But are there conditions that can be put on $Y$ to make it so? $\endgroup$ – Dr Shello May 17 '11 at 6:05

As others have pointed out, the generic case (whatever that should mean in this case) is that the loop structure on a loop space is not unique. However, things get quite interesting whenever we have a space that actually does have a unique loop structure. I highly recommend looking at:

Dwyer, Miller, Wilkerson: The homotopic uniqueness of $BS^3$, LNM 1298


Dwyer, Miller, Wilkerson: Homotopical uniqueness of classifying spaces. Topology 31 (1992), no. 1, 29–45.


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