TITLE: Constructing a odd homeomorphism between $A$ and $S^n$.
QUESTION [5 upvotes]: Let $A\subset\mathbb{R}^N\setminus\{0\}$ be a closed symmetric set ($x\in A$ then $-x\in A$). Suppose that $A$ is homeomorphic to some sphere $S^n$, $n\leq N$ ($n$ is the dimension of the sphere). Is it possible to construct a homeomorphism $F:A\to S^n$ such that $F$ is odd?
Update: this problem was solved here.

REPLY [3 votes]: Here's an idea, but I don't see how to finish it. You have a homeomorphism $f\colon A\to S^n$, and this induces an involution $\alpha\colon S^n\to S^n$ given by $f(x)\mapsto f(-x)$. Using a variant of Alexander's trick, one can show that $\pi_0({\rm Homeo}(S^n))\cong \mathbb Z_2$, with the two classes represented by orientation-reversing and orientation-preserving homeomorphisms, respectively. Therefore $\alpha$ is isotopic to the antipodal map $x\mapsto -x$ provided that it does the same thing to orientation as the antipodal map. Notice that $\alpha$ is fixed-point free. So the Lefschetz trace must be zero. In this simple case of spheres, the Lefschetz trace of a homeomorphism $f$ is $1+(-1)^n\deg f$. So $\deg \alpha=(-1)^{n+1}$, which is the same as the degree of the antipodal map. Therefore $\alpha$ is isotopic to the antipodal map. If $\alpha$ were ambient isotopic to the antipodal map we'd be done, since an ambient isotopy $H\colon S^1\times I\to S^n$ would be a function where $H(x,0)=x$ and $H(\alpha(x),1)=-x$. So letting $g(x)=H(x,1)$ the map $gf\colon A\to S^n$ would commute withe the antipodal map, as desired. Thus we are left with the problem of promoting an isotopy to an ambient isotopy, which is not always possible. Indeed, Alexander's trick creates nasty singularities, which might mean that it can't be done in general. If you switch to ${\rm Diff}(S^n)$ rather than ${\rm Homeo}(S^n)$, you could probably assume that isotopies are promotable to ambient isotopies, but $\pi_0({\rm Diff}(S^n))$ would become more complicated. Indeed, for $n\geq 5$, $\pi_0({\rm Diff}^+(S^n))\cong\Theta_{n+1}$, the group of exotic $(n+1)$-spheres. So if you're going to find a counterexample, $n+1$ will have to be at least $7$, when the first exotic spheres occur.