Timeline for Do all spaces doubly covered by $S^{2n}$ have the homeomorphism type of $\mathbb{P}^{2n}_{\mathbb{R}}$?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Oct 30, 2021 at 17:13 | vote | accept | Rafi | ||
| Oct 30, 2021 at 12:25 | comment | added | Igor Belegradek | @RafayAshary: what you stated is correct. Also in this case any homotopy equivalence is simple because the Whitehead group of $\mathbb Z_2$ is trivial. | |
| Oct 30, 2021 at 8:42 | comment | added | Rafi | Awesome, thanks! So if I'm understanding correctly, the idea is that for $n\geq 3$, there is a large ($\geq 2$) number of homeomorphism classes of manifolds simply homotopy equivalent to real projective space of dimension $2n$, and the Poincaré conjecture for topological manifolds of dimension $2n$ implies that any of these have a (universally covering) double cover homeomorphic to $S^{2n}$, together demonstrating the existence of the desired counterexample(s)? | |
| Oct 30, 2021 at 2:46 | history | edited | Igor Belegradek | CC BY-SA 4.0 |
added 2 characters in body
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| Oct 30, 2021 at 2:36 | history | answered | Igor Belegradek | CC BY-SA 4.0 |