If $\Omega X \simeq \Omega Y$ then is $X \simeq Y$ for $X,Y$ simply connected?

Assuming $X,Y$ are nice spaces like CW of course.

Clearly this is true by Whitehead, but I am looking for a more enlightening proof.

  • 7
    $\begingroup$ As written, this is false. What follows from Whitehead is that given a map $f: X\to Y$ between connected spaces, which induces an equivalence on loop spaces, it was an equivalence to begin with. This follows for example also from the fact that loop space and the bar construction constitute inverse equivalences between connected spaces and grouplike $\mathbb{E} _1$-spaces. $\endgroup$ Nov 22, 2019 at 13:14

1 Answer 1


To expand on my comment: As written (i.e. without requiring a map $f:X\to Y$), this is false in general. For an example, write $S^2$ as homogeneous space, $S^2=SU(2)/U(1)$. This exhibits $S^2$ as the homotopy fiber of a map $$ BU(1)\to BSU(2). $$ The space $BU(1)\times SU(2) \simeq \mathbb{C}P^\infty \times S^3$ is also the homotopy fiber of a map $BU(1)\to BSU(2)$, namely the constant map.

$S^2$ and $\mathbb{C}P^\infty\times S^3$ are obviously not homotopy equivalent. But after looping once, both our maps $BU(1)\to BSU(2)$ turn into maps $U(1)\to SU(2)$, i.e. $S^1\to S^3$, and thus become homotopic. So $$ \Omega S^2 \simeq \Omega (\mathbb{C}P^\infty \times S^3). $$

  • $\begingroup$ That is a really nice counterexample. Thank you! $\endgroup$
    – WhoAmI
    Nov 22, 2019 at 13:58

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