Timeline for Path-lifting property: function space interpretation
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 7, 2018 at 2:34 | comment | added | Hempelicious | Thank you for your answer! I think I fully understand how this works now. Do you want to post a similar answer on math.SE? I could also post my understanding of your answer there. Again, thanks! | |
| Jun 7, 2018 at 2:33 | vote | accept | Hempelicious | ||
| May 31, 2018 at 23:23 | comment | added | Jeremy Brazas | You don't have to exclude $n=1$...I guess it's a preference. But the argument absolutely requires the form above for $n\geq 2$. Why would you want to consider $n=1$ as something special when you don't have to? My instinct to start with $n\geq 3$ is related to covers of simplicial complexes...that is all. | |
| May 31, 2018 at 18:52 | comment | added | Steve D | Right, I think I followed the argument, but then seeing this requirement made me second-guess it. Sure, we can always refine, but my question was more "why do we have to?" | |
| May 31, 2018 at 18:38 | comment | added | Jeremy Brazas | @SteveD Well, $n= 2$ could be included for this particular proof but this is not really an important feature. You could always take finer partitions and consider only sets of the form given where $n$ is greater than some fixed number. Allowing $n=1$ would just require an unnecessary extra case to write. | |
| May 31, 2018 at 17:57 | comment | added | Steve D | Why do we need $n\ge3$? | |
| May 29, 2018 at 21:49 | history | edited | Jeremy Brazas | CC BY-SA 4.0 |
added 52 characters in body
|
| May 29, 2018 at 19:59 | history | answered | Jeremy Brazas | CC BY-SA 4.0 |