Timeline for Standard proof that cyclic ordering of edges is preserved under planar graph homotopy?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 9, 2016 at 19:50 | comment | added | Jan Kyncl | The "only if" direction also fails if the graph itself is disconnected and has at least one cycle. | |
| Sep 9, 2016 at 10:12 | comment | added | Allen Hatcher | For examples where "only if" fails, start with a graph in ${\mathbb R}^2$ whose complement is not connected, then add a point at infinity to get a graph in $S^2={\mathbb R}^2\cup\{\infty\}$. For each complementary component of this graph in $S^2$, if one deletes a point from this component, one obtains an embedding of the graph in ${\mathbb R}^2$ with the same cyclic orderings at vertices, but the various embeddings obtained this way are not homotopic in the sense defined in the question. (This notion of "homotopy" is usually called "isotopy".) | |
| Sep 9, 2016 at 7:34 | history | edited | sk1 | CC BY-SA 3.0 |
Based on initial comments, rewrote the question for clarity and future usefulness.
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| Sep 9, 2016 at 6:30 | comment | added | Ilya Bogdanov | @Allen: Do you have a counterexample for the converse? I would say that "if and only if" is fine... | |
| Sep 9, 2016 at 1:27 | comment | added | Allen Hatcher | In the statement of the theorem, the "if and only if" should be just "if". | |
| Sep 9, 2016 at 0:26 | review | First posts | |||
| Sep 9, 2016 at 1:06 | |||||
| Sep 9, 2016 at 0:23 | history | asked | sk1 | CC BY-SA 3.0 |