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Apr 4, 2011 at 8:56 comment added Sergey Melikhov Thanks for interest, but I think you're solving a different problem. Your notation $N/G$ suggests that you're assuming that $N$ is $G$-invariant, but it is not (cf. Tom Goodwillie's comment above). OK, let's interpret $N/G$ as the image of $N$ under the quotient map. In my example with the antipodal involution on the sphere, $N\subset N'$ would imply $N=N'$ and has nothing to do with $N/G\subset N/G$, which is automatic. Your isotopy $\bar F$ would be the identity, etc.
Apr 4, 2011 at 2:45 history answered Qayum Khan CC BY-SA 2.5