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Dec 23, 2019 at 19:37 comment added Paul Cusson Yes I meant $+2\pi /n$. And thanks for clarifying about the ping, I wasn't sure how that worked exactly. By you are right, mean about my counterexample, right? Thanks for the input.
Dec 23, 2019 at 19:32 vote accept Paul Cusson
Dec 23, 2019 at 19:28 comment added Mark Grant PS I got pinged anyway, the author of any question or answer always gets notified of any comments on their post.
Dec 23, 2019 at 19:27 comment added Mark Grant About your second question, there has been some work done on extending Borsuk-Ulam to homology spheres, see the references in the linked thesis (Conner-Floyd, Munkholm,...)
Dec 23, 2019 at 19:26 comment added Mark Grant @PaulCusson: You are right, I think, and your example above has $A(g)$ empty whilst having points for which $g(x)=g(f(x))$. (I suppose you meant to write $f(\theta)=\theta + 2\pi/n$ or something similar?) There doesn't seem to be much in the literature about your precise question.
Dec 23, 2019 at 0:25 comment added Paul Cusson Sorry, @Mark I forgot to ping you
Dec 22, 2019 at 21:08 comment added Paul Cusson In particular, $g : S^1 \to \mathbb{R}$ such that $g(\theta) = \cos(\theta)$, and $f:S^1 \to S^1$ such that $f(\theta) = \theta + 1/n$ for some natural $n$ large enough. The map $f$ is such that $f^n$ is the identity on $S^1$, and we should get $A(g)$ empty, but $g(S^1)$ and $g(f(S^1))$ have two points of intersections, so two such "fixed points".
Dec 21, 2019 at 6:02 comment added Paul Cusson This is interesting. But what if I don't want other powers of $f$ to fix the value, and just want $g(x)= g(f(x))$? We could still have $A(g)$ empty in this case, if for every $x$ some other power doesn't fix the value.
Dec 20, 2019 at 12:08 history answered Mark Grant CC BY-SA 4.0