Let $X_1$ and $X_2$ be CW-complexes such that their loop spaces are homotopy equivalent, $\Omega X_1 \simeq \Omega X_2$. I wonder why the spaces themselves are homotopy equivalent, $X_1 \simeq X_2$. I tried using the adjunction with the suspension $\Sigma$ and Hurewicz theorem that a map inducing isomorphism on homotopy groups is a homotopy equivalence, but then one needs a map $X_1 \to X_2$ inducing homotopy equivalence on loop spaces. Where am I wrong?
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2$\begingroup$ To recover the space, you need, first of all, to know that $X_1$ and $X_2$ are path-connected to avoid trivial counterexamples. Then, you need to have $\Omega X_1 \cong \Omega X_2$ as $A_\infty$ algebras, not just as spaces. $\endgroup$ – Justin Young Nov 25 '16 at 16:13
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No. The easiest place to see this (and a place you can find a plethora of examples) is with classifying spaces of groups. Of course $\Omega BG \simeq G$. But $BO(2)$ and $B(S^1 \times \Bbb Z/2) = \Bbb{CP}^\infty \times \Bbb{RP}^\infty$ are not homotopy equivalent, despite the underlying groups being homeomorphic. To see this, check the rational cohomology. There is a fiber sequence $\Bbb{RP}^2 \to BO(2) \to BSO(3)$, and because the fiber is rationally trivial, $BO(2) \to BSO(3)$ is a rational equivalence. Further we have the fiber sequence $\Bbb{RP}^\infty \to BSU(2) \to BSO(3)$, and $BSU(2) = \Bbb{HP}^\infty$. So $H^*(BO(2);\Bbb Q) = H^*(BSU(2);\Bbb Q) = \Bbb Q[u]$, $|u| = 4$. The rational cohomology of the other space is polynomial on a degree 2 generator.
