Timeline for Is $\mathbb R^3$ the square of some topological space?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 6, 2020 at 18:21 | comment | added | Matthias Wendt | This argument is also given as exercise in Hatcher's "More exercises in algebraic topology". | |
| Oct 4, 2020 at 14:39 | history | edited | Henno Brandsma | CC BY-SA 4.0 |
Added proof from link to get more standalone post.
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| Apr 5, 2011 at 5:42 | comment | added | Anton Geraschenko | @Martin: The homeomorphism $(X\times X)\times (X\times X)\cong \mathbb R^3 \times \mathbb R^3$ respects projections by construction, so swapping the "two factors" (which I've emphasized with parentheses) on the left hand side corresponds to swapping the two factors on the right hand side. | |
| Apr 4, 2011 at 15:05 | comment | added | Martin Brandenburg | I don't understand this step in the proof: Why does the map $X^4 \to X^4, (a,b,c,d) \mapsto (c,d,a,b)$ correspond to the map $R^6 \to R^6, (p,q,r,s,t,u) \mapsto (s,t,u,p,q,r)$? I mean, the homeomorphism is not supposed to commute with projections ... | |
| Apr 4, 2011 at 5:16 | comment | added | Yaakov Baruch | Quoting from the link: "The paper also refers to an earlier paper ("The cartesian product of a certain nonmanifold and a line is E4", R.H. Bing, Annals of Mathematics series 2 vol 70 1959 pp. 399–412) which constructs an extremely pathological space B, called the "dogbone space", not even a manifold, which nevertheless has B × R^3 = R4." This is relevant to my comment to the OP. | |
| Apr 4, 2011 at 2:24 | comment | added | Richard Dore | I hope no one misses this nice alternative proof because it's behind a link. | |
| Apr 2, 2011 at 21:25 | history | answered | Henno Brandsma | CC BY-SA 2.5 |