Number of antipodal pairs of $n$-sphere points mapped to discrete points in $\mathbb{R}^n$ I'm thinking on the Borsuk-Ulam theorem, which states that

If $f\colon\mathbb{S}^n\to\mathbb{R}^n$ is continuous, then there exists an $x$ such that $f(x)=f(-x)$.

This means that if $f$ is continuous, then at least one pair of antipodal points of the $n$-sphere are mapped to the same point in $\mathbb{R}^n$.
Now, I'm wondering if a continuous map $f$ can be constructed so as a fixed number of antipodal pairs are maps to the same fixed number of points in $\mathbb{R}^n$.
 A: Take polar coordinates, with latitude denoted by $\phi$ and longitude by $\theta$, so that
$$
S(\phi, \theta) = \pmatrix{
\cos \phi \cos \theta \\
\cos \phi \sin \theta \\
\sin \phi}
$$
and $\phi$ ranges from $-\pi/2$ to $\pi/2$, and do everything $\bmod 2\pi$. Consider the map $C$ from the unit sphere to the plane given by $(x, y, z) \mapsto (x, z)$. This preserves latitudes (i.e., two points with the same latitude end up with the same second coordinate), so the only possible antipodal pairs are on the equator $\phi = 0$. Antipodal pairs, in polar coords, are $p_\theta = (0, \theta)$ and $p'_\theta= (0, \theta + \pi)$. The $x$-coordinates of these points are $\cos \theta$ and $\cos (\theta + \pi) = -\cos \theta$. For these to be equal when projected, i.e., for
$$
C(S(p_\theta)) = C(S(p'_\theta))
$$
therefore requires that $\cos\theta = 0$. So the particular map $C$ from $\Bbb S^2$ to $\Bbb R^2$ has exactly one antipodal pair that's mapped to a single point; let's call that a "good pair".
Now consider the map $M_2$ (for "multiply by two") from $\Bbb S^2 \to \Bbb S^2$ defined by $(\phi, \theta) \mapsto (\phi, 2\theta)$ in polar coordinates. Let's define
$$
C_2: \Bbb S^2 \to \Bbb R^2 : (\phi, \theta) \mapsto C(S(M_2(\phi, \theta))).
$$
For a point $X = (\phi, \theta)$ to be an element of a good pair for $C_2$ requires that $S(M_2(X))$ be an element of a good pair for $C$, so that $M_2(X)$ must have $\phi = 0$ and $\theta = 0, \pi$, so $\theta$ must be $0, \pi/2, \pi, 3\pi/2$. So $C_2$ has exactly two good pairs.
If you defined $M_k (\phi, \theta) = (\phi, k\theta)$, you can define $C_k$ analogously and get yourself exactly $k$ good pairs, so your question, for $n = 2$, is answered in the affirmative.
Hell. I now realize I should have done this for $\Bbb S^1$ first, because then I could observe that the map $C$ is just the suspension of the map $c: \Bbb S^1 \to \Bbb  S^1:(x, y) \mapsto $x$.
But that leads to the next observation, which is that by suspension of $C$, you get a similar map from $\Bbb S^3$ to $\Bbb R^3$, and so on. So the answer is affirmative in all dimensions.
N.B.: Because $M_k$ produces a map with $k$ good pairs, you might think "Well maybe if I look at $M_0$, I'll get zero antipodal good pairs, and I'll have a contradiction to Borsuk-Ulam!" Alas, that's not the case. When you try $M = 0$, it turns out that all equatorial pairs are mapped to the origin, so you have infinitely many good pairs rather than zero.
