Timeline for Dimension of a set detected by a homology class
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 19, 2014 at 18:59 | comment | added | Anton Petrunin | @IgorBelegradek, All this follows directly from the construction. | |
| Apr 19, 2014 at 18:54 | comment | added | Igor Belegradek | I could not even find where a)-b) are stated (let alone proved). Appparently, pseudo-circle is not homogeneous, so homogeneity cannot be used to show b). | |
| Apr 19, 2014 at 18:46 | comment | added | Anton Petrunin | @IgorBelegradek, I think all this standard, isn't it? I only wanted to indicate the construction. | |
| Apr 19, 2014 at 18:37 | comment | added | Igor Belegradek | This is a nice picture but I'd rather see an explanation why the pseudo-circle has a) infinite cyclic first Cech cohomology and b) no manifold point. | |
| Apr 19, 2014 at 17:41 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
+pic
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| Apr 18, 2014 at 22:17 | comment | added | Włodzimierz Holsztyński | @Igor Belegradek (quote: search the web on "pseudo-circle" and "continuum")--no need to. But thank you for your effort. | |
| Apr 18, 2014 at 21:02 | comment | added | Igor Belegradek | I gather, Cech cohomology of the pseudo-circle are the same as for the circle, so by Alexander duality its complement in the plane has two components and has first homology isomorphic to $\mathbb Z$. By the classification of surfaces this means that the complement's components are open disk and open annulus. Joining them by a handle gives the desired embedding into the $2$-torus. The only thing I am not sure about is whether the first Cech cohomology is $\mathbb Z$, but if this weren't true, why would it be called pseudo-circle? | |
| Apr 18, 2014 at 19:43 | comment | added | Igor Belegradek | @WlodzimierzHolsztynski: search the web on "pseudo-circle" and "continuum". | |
| Apr 18, 2014 at 18:22 | comment | added | Włodzimierz Holsztyński | Anton, your link says "pseudo-circle", while it points to wikipedia article on pseudo-arc, and it does not mention pseudo-circle at all. Could you comment about it? | |
| Apr 18, 2014 at 18:18 | comment | added | Włodzimierz Holsztyński | Anton, what do you mean by statement: For sure you can not expect a subset homeomorphic to $\mathbb R^p$ ? | |
| Apr 18, 2014 at 16:25 | vote | accept | Tom Goodwillie | ||
| Apr 18, 2014 at 16:25 | comment | added | Tom Goodwillie | Thank you, Anton. I did not know that there were such strange examples in such low dimensions. | |
| Apr 18, 2014 at 14:54 | comment | added | Igor Belegradek | Why is the complement homeomorphic to a cylinder? | |
| Apr 18, 2014 at 13:38 | history | answered | Anton Petrunin | CC BY-SA 3.0 |