Timeline for Are loop spaces of homotopically equivalent spaces homotopically equivalent?
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| Mar 10, 2013 at 23:33 | comment | added | Ricardo Andrade | (continuation) Since $(\Omega_{f(x)} g)\circ(\Omega_x f)$ is a homotopy equivalence, we conclude that $\Omega_{f(x)} g$ has a homotopy right inverse. Analogously, $\Omega_{f(x)} g$ has a homotopy left inverse because $(\Omega_{g(f(x))} f)\circ(\Omega_{f(x)} g)$ is a homotopy equivalence. Consequently, $\Omega_{f(x)}g$ is a homotopy equivalence. Finally, since $(\Omega_{f(x)} g)\circ(\Omega_x f)$ and $\Omega_{f(x)} g$ are both homotopy equivalences, so is $\Omega_x f$. This argument is a topological version of the usual proof that homotopy equivalences induce isomorphisms on fundamental groups. | |
| Mar 10, 2013 at 23:23 | comment | added | Ricardo Andrade | Actually, it is also straightforward to show that the map induced by any homotopy equivalence $f:X\to Y$ on based loop spaces is a homotopy equivalence. Simply observe that if $g:Y\to X$ is a homotopy inverse to $f$, then $(\Omega_{f(x)} g) \circ(\Omega_x f)$ is homotopic to "conjugation" by the path $H(x,-)$ for any homotopy $H:\mathrm{id}_X \simeq g\circ f$. Moreover, this "conjugation" map is a homotopy equivalence, since conjugation by the reverse path is a homotopy inverse. Similarly, $(\Omega_{g(f(x))} f) \circ(\Omega_{f(x)} g)$ is a homotopy equivalence. (to be continued) | |
| Mar 10, 2013 at 18:53 | comment | added | Hiro | Thanks for your comment. I wonder how one can prove that $\Omega (F_{t})$ is actually a homotopy. Is it always continuous without any assumption such as locally compactness? | |
| Mar 10, 2013 at 18:43 | history | answered | Mark Grant | CC BY-SA 3.0 |