In the setting of the Borsuk-Ulam theorem, is it true that for any distance $d$ there exist two points $A$ and $B$ with distance $d$, such that $f(A)=f(B)$? (Two antipodal points having the maximal distance).
A generalization of the Borsuk-Ulam theorem?
general-topology
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The diagram is a bit confusing since he didn't animate $A$ and $B$ as he did in the $\mathbb{R}^2$ case, but the argument is the same. To restate the problem for the sphere: for any two continuous real functions $f, g$ on the sphere, there exists a pair of antipodal points $x, -x$ where $f(x) = f(-x)$ and $g(x) = g(-x)$. In this case we're using temperature and pressure as the two continuous functions.
Let $P$ be any continuous path on the sphere from $A$ to $B$, and let $P'$ be the set of antipodal points to $P$. If take both $P$ and $P'$ together, we get a continuous loop that goes from $A$ to $B$ and back. In essence this is the same as he did for $\mathbb{R}^2$, only in that case he chose a very specific $P$ in order to get the equator.
Now for any $P$ and its $P'$, we can consider the temperature of $A$ and $B$ as they move along the path, across from each other. If the temperatures at the starting points are the same, we've found a same-temperature point pair and we're done. Otherwise, one of these is greater (let's say $A$, but the argument can be reversed). We can make the same "graph of temperature" that he did for $\mathbb{R}^2$ and observe the same phenomenon. When $A$ moves from its initial position to $B$'s initial position, it must drop from its initial temperature until it reaches $B$'s initial temperature. $B$ likewise ends at $A$'s initial position, so must rise from its initial temperature until it reaches $A$'s initial temperature. It follows that there has to be a point where the two temperature lines cross (i.e. the temperatures are the same).
The second part of the argument says that at least one of these same-temperature pairs have the same pressure. He argues they must form a set ("club") which separate the initial positions of $A$ and $B$ (any path from $A$ to $B$ must end up passing through the set). To prove this, assume that that there is a gap in the set that lets us thread a path from $A$ to $B$ without touching the set. We know from the previous part that this path must have a same-temperature antipodal pair, so this point pair should be in the set. This is a contradiction of the path we chose, so it must be that no such gap exists. Some continuous separator must exist, which splits $A$ and $B$.
Let's call this continuous separator set $S$. Remember, $S$ is made of antipodal point pairs, so let's select an arbitrary pair to be $C, D$. Since $S$ is connected we can always find a path $P$ from $C$ to $D$ and take its antipodal half $P'$ (also from $S$) to make a continuous loop of antipodal points. We reuse the same argument as we did for temperature in the first half (only now for pressure) to assert that there must be an antipodal point pair in $S$ with equal pressure. Since all antipodal point pairs of $S$ have equal temperature, this point pair is the one we're looking for.
I think that the following should work:
If there were some $f:S^2 \to \mathbb R^2$, $f(x) \neq f(-x)$, then $f(x)-f(-x) \neq 0$ for any $x \in S^2$.
This gives a function $\phi:S^2 \to S^1$ given by $x \mapsto \frac{f(x)-f(-x)}{\|f(x)-f(-x)\|}$.
Notice that $\phi(x) = - \phi(-x)$ for all $x \in S^2$.
Compose this map with the double covering self-map $\rho : S^1 \to S^1$ which makes the identification $\rho(z)=\rho(-z)$ for all $z \in S^1$.
Now, $\rho \circ \phi(x)=\rho \circ \phi(-x)$.
By the universal property of the quotient map, this induces a map $$\widetilde{\rho \circ \phi}: \mathbb RP^2 \to (S^1 \cong \mathbb RP^1) $$ To get the final contradiction, this map induces a nontrivial homomorphism from the group $\mathbb Z / 2 \mathbb Z = \pi_1(\mathbb RP^2)$ to the group $\mathbb Z = \pi_1(\mathbb S^1)$, because a closed path in $\mathbb RP^2$ representing a generator of $\pi_1(\mathbb RP^2)$ lifts to a path in $S^2$ with antipodal endpoints, which maps to a path in $S^1$ with antipodal endpoints, which projects via the map $\rho$ to a closed path in $S^1$ representing an odd element of $\pi_1(S^1) = \mathbb Z$.
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Remark: As pointed out by Thomas Anton, the arguments below do not make sense: $f(u(t))$ is an element of $\mathbb R^n$, not an element of $S^n$. Thus $B = f(u(t_r)) \notin S^n$. And even if $B$ were in $S^n$, there is no reason to expect that $f(A) = f(B)$.
I tried to delete my answer, but this is impossible for accepted answers.
Here is the original answer:
Yes. The Borsuk-Ulam theorem states that for any continuous map $f : S^n \to \mathbb R^n$ there exists $x \in S^n$ such that $f(x) = f(-x)$. The points $x, y = -x$ have the distance $\lVert y - x \rVert = 2$. Now let $u : [0,1] \to S^n$ be any path from $x$ to $y$. The function $\phi : [0,1] \to \mathbb R, \phi(t) = \lVert f(u(t)) - x \rVert$, is continuous and satisfies $\phi(0) = 0, \phi(1) = 2$. By the intermediate value theorem for each $r \in [0,2]$ there exists $t_r \in [0,1]$ such that $\phi(t_r) = r$. Now take $A = x, B = f(u(t_r))$.
By the way, this proof works for any metric $d$ on $S^n$ which generates the standard topology.