Timeline for Wild half-line in a Euclidean space
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 28, 2016 at 14:33 | vote | accept | Anton Petrunin | ||
| Jan 25, 2016 at 20:01 | answer | added | Anton Petrunin | timeline score: 4 | |
| Jan 25, 2016 at 19:05 | comment | added | YCor | I can believe so, but I haven't grasped the argument. | |
| Jan 25, 2016 at 18:45 | comment | added | Anton Petrunin | @YCor, but the answer is the same. | |
| Jan 25, 2016 at 17:54 | comment | added | YCor | I was asking about an injective path, not an injective loop. | |
| Jan 25, 2016 at 17:09 | comment | added | Anton Petrunin | @YCor it is a 1-dimensional subcomplex in the triangulation coming from the suspension, but (by obvious reason) it is not subcomplex in the standard triangulation of sphere. | |
| Jan 25, 2016 at 14:57 | comment | added | YCor | So, I don't understand your comment. The "double-suspension-equator" is not an injective combinatorial path, is it? | |
| Jan 25, 2016 at 12:58 | comment | added | Anton Petrunin | @YCor double suspension over $X$ is the joint $\mathbb{S}^1*X$ and $\mathbb{S}^1$ is its equator. | |
| Jan 25, 2016 at 12:49 | comment | added | YCor | What is "double suspension equator"? | |
| Jan 25, 2016 at 11:33 | comment | added | Anton Petrunin | @YCor The double suspension over Poincaré sphere is homeomorphic to $\mathbb{S}^5$. The complement of the double-suspension-equator in it is not simply connected. | |
| Jan 25, 2016 at 11:27 | comment | added | YCor | Subquestion: does there exist a simplicial complex, homeomorphic to some sphere, and in which there exists an injective combinatorial path (= some consecutive edge with no return) whose complement is not simply connected? | |
| Jan 25, 2016 at 11:25 | comment | added | YCor | In the comment you might say that the embedding of the half-line is proper . | |
| Jan 25, 2016 at 10:26 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Jan 24, 2016 at 23:27 | comment | added | YCor | OK... So passing to the 1-point compactification, it yields a sphere in which the complement of some segment is not simply connected. | |
| Jan 24, 2016 at 23:24 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
added 1 character in body; edited title
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| Jan 24, 2016 at 23:21 | comment | added | Anton Petrunin | @YCor the cone is infinite. | |
| Jan 24, 2016 at 23:17 | comment | added | YCor | Btw do you define the cone of $X$ from $X\times [0,1]$ by crushing $X\times\{0\}$ to a point, or from $X\times [0,+\infty\mathclose[$? | |
| Jan 24, 2016 at 22:45 | history | asked | Anton Petrunin | CC BY-SA 3.0 |