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I've come upon this MO post, and I was wondering if there is a way to generalize what is said in the answer at the very bottom. Specifically, we have that given continuous maps $f: \mathbb{S}^n \to \mathbb{S}^n$, and $g: \mathbb{S}^n \to \mathbb{R}^n$, if $f$ is an involution, then there is a point $x \in \mathbb{S}^n$ such that $g(x) = g(f(x))$, i.e. the value at $x$ is fixed under the transformation of the sphere by $f$.

We know from the answer of the post that for a general $f$, there need not be such a fixed point. But now what if instead $f$ is such that there exists a $k \in \mathbb{N}$ where $f^k$ is the identity map, do we still get such a fixed point? Furthermore, do we know about other spaces besides the sphere that have this property when $f$ is an involution?

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A first observation is that a map $f:S^n\to S^n$ with $f^k=\operatorname{Id}_{S^n}$ generates a topological action of the cyclic group $\mathbb{Z}/k$ on $S^n$. So we are talking about generalising from $\mathbb{Z}/2$ actions to $\mathbb{Z}/k$ actions.

There are many generalisations of the Borsuk-Ulam theorem of this kind in the literature. Most of them concern estimates of the dimension of the set $$ A(g)=\{x\in S^n \mid g(x)=g(f^i(x))\mbox{ for }i=1,\ldots , k\}, $$ for various flavours of map $g:S^n\to M^m$.

Interpreted this way, the answer to your first question appears to be negative in general, by a theorem of Munkholm in

Munkholm, H. J., On the Borsuk-Ulam theorem for (Z_{p^ a}) actions on (S^{2n-1}) and maps (S^{2n-1} \to R^ m), Osaka J. Math. 7, 451-456 (1970). ZBL0211.55701.

In particular, let $k=p^a\neq 9$ be an odd prime power with $a>1$, and let $\omega=\operatorname{exp}(2\pi i/p^a)$ be a primitive $p^a$-th root of unity. Let $f:S^{2n-1}\to S^{2n-1}$ be given in complex coordinates by $$ f(z_1,\ldots , z_n)=(\omega z_1,\ldots ,\omega z_n). $$ Then Munkholm's "mod $p^a$ Borsuk-Ulam anti-theorem" seems to imply that there exists a continuous map $g:S^{2n-1}\to \mathbb{R}^{2n-1}$ with $A(g)=\emptyset$.

For a summary of positive results in the literature, you could do worse than Chapter 1 of Yuri Turygin's thesis at the Unviersity of Florida.

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  • $\begingroup$ 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. $\endgroup$
    – Paul Cusson
    Commented Dec 21, 2019 at 6:02
  • $\begingroup$ 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". $\endgroup$
    – Paul Cusson
    Commented Dec 22, 2019 at 21:08
  • $\begingroup$ Sorry, @Mark I forgot to ping you $\endgroup$
    – Paul Cusson
    Commented Dec 23, 2019 at 0:25
  • $\begingroup$ @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. $\endgroup$
    – Mark Grant
    Commented Dec 23, 2019 at 19:26
  • $\begingroup$ 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,...) $\endgroup$
    – Mark Grant
    Commented Dec 23, 2019 at 19:27

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