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.
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.
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 $$ \Omega S^2 \simeq \Omega (\mathbb{C}P^\infty \times S^3). $$