Timeline for Universal cover of the configuration space of points on surface
Current License: CC BY-SA 4.0
6 events
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| Nov 17, 2023 at 11:53 | comment | added | Roman | @MarkGrant, Yes, the evaluation map ${\rm Homeo_0}(S)/{\rm Homeo_0}(S, V) \rightarrow C(S, n)$ is a homeomorphism. The subtlest part is the openness of this map. The standard proof of the openness seems also to extend to the openness of the map in my question. I think, I can then to do an inductive argument to show that my map is a bijection, but it is a bit annoying. I wonder if nobody came to this before, or if there is a simpler proof. | |
| Nov 17, 2023 at 11:49 | comment | added | Roman | @alesia, I'd love to see such an argument, but unfortunately, I couldn't so far produce it myself. Please note that all the homeomorphisms in question are supposed to be isotopic to identity on $S$. | |
| Nov 17, 2023 at 7:06 | comment | added | Mark Grant | There is an "evaluation at $V$" map from $\mathrm{Homeo}_0(S)$ to $C(S,n)$, which I believe is a fibration with fibre $\mathrm{Homeo}_0(S,V)$. Does this imply that the quotient is in fact $C(S,n)$? | |
| Nov 16, 2023 at 22:27 | comment | added | alesia | doesn't this follow shortly enough from the isomorphism between braid groups and mapping class groups? | |
| Nov 16, 2023 at 22:11 | history | edited | Roman | CC BY-SA 4.0 |
added 46 characters in body
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| Nov 16, 2023 at 20:57 | history | asked | Roman | CC BY-SA 4.0 |