A question on Borsuk–Ulam theorem when $\Bbb S^n$ viewed as topological sphere. Borsuk–Ulam theorem states that:

For every continuous function $f:\Bbb S^n\to \Bbb R^n$ maps some pair of antipodal points to the same point. i.e. there exists an $x\in\Bbb S^n$ such that $f(x)=f(-x)$.

Is this true for every  topological sphere? in this case what is the definition of antipodal points? 
 A: Let $X$ be a topological space, homeomorphic to the $n$-sphere $\Bbb{S}^n$ via a homeomorphism $h:\Bbb{S}^n\to X$. Let $g:X \to \mathbb{R}^n$ be a continuous function. Then $g\circ h:\Bbb{S}^n\to\Bbb{R}^n$ is continuous, and so it satisfies the Borsuk-Ulam theorem. Hence there exists an $x\in\Bbb{S}^n$ such that $(g\circ h)(x)=(g\circ h)(-x)$.
From the point of view of $X$, there exists points $h(x)$ and $h(-x)$ in $X$ such that $g(h(x))=g(h(-x))$. So the points $h(x)$ and $h(-x)$ take the place of our antipodal points.
A: It’s in itself pretty specific for the involution map $i(x) = -x$. (an involution $j: X \to X$ satisfies $j(j(x)) = x$, and the negation map we also have it has no fixed points on the sphere.
But if we have a space $X$ with a homeomorphism $h: X \to \mathbb{S}^n$, then we can define $j : X \to X$ by $j(x) = h^{-1}(i(h(x)))$, and note that $j$ has no fixed points , as 
$$j(x) =x \Leftrightarrow h(j(x)) =  h(x) \Leftrightarrow i(h(x)) = h(x)\text{.}$$
Also, if $f: X \to \mathbb{R}^n$ is continuous, so is $f \circ h^{-1}: \mathbb{S}^n \to \mathbb{R}^n$ so classical Borsuk-Ulam says there is some $p \in \mathbb{S}^n$ with $f(h^{-1}(-p) = f(h^{-1}(p))$ but then for $p’ = h^{-1}(p)$ we have 
$$f(j(p’)) = f(h^{-1}(i(h(p’))) = f(h^{-1}(-h(p’))) = f(h^{-1}(-p) = f(h^{-1}(p)) = f(p’)$$ unpacking all definitions and givens.  
So for every topological sphere $X$ there is a fix-point free involution $j: X \to X$ such that for every continuous $f: X \to \mathbb{R}^n$ there is $p \in X$ such that $f(j(p)) = f(p)$, similarly to what we have for $i$ on the classical sphere. I think that is the best we can do. We cannot expect the same $i$ to work, but every topological sphere has an analogue map.
