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Feb 24, 2020 at 5:53 comment added Meths Got it, thanks for your help!
Feb 24, 2020 at 3:56 comment added Ty Ghaswala It's not immediately obvious, but It does induce a continuous map between unordered configuration spaces. One way to see this is to first check it on the ordered configuration spaces, and then make sure the induced inclusion respects the quotient maps. However, it's clearer if you use the viewpoint you presented in your original question, of looking at elements of the surface braid group as a collection of paths on $D$ or $M$. Then a collection of paths in $D$ is clearly a collection of paths in $M$ once you include $D$ into $M$.
Feb 22, 2020 at 11:45 comment added Meths Bump... quick question: for your (induced homomorphism) argument to be true, doesn't this also mean there has to be a continuous map from the (unordered) configuration space of $D$ to that of $D\subset M$? That doesn't seem immediately obvious to me. Sorry for the posthumous enquiry!
Jan 24, 2020 at 14:28 vote accept Meths
Jan 24, 2020 at 14:17 comment added Meths Simple enough !
Jan 24, 2020 at 1:37 history answered Ty Ghaswala CC BY-SA 4.0