Intuition behind Borsuk-Ulam Theorem

Question

I watched the following video to get more intuition behind Borsuk-Ulam Theorem.

The first part of the video was very clear for me, as I understood it considers only $R^2$ dimension and points $A$ and $B$ moving along the equator and during the video we track the temperature of point $A$ and $B$ along the equator.

The following is the picture from the second part.

In the second part $R^3$ is considered, and instead of tracking the temperatures along the equator, we track the temperature along the arbitrary path from $A$ to $B$ along the sphere, but along this part we don't move $A$ and $B$ there is no intersection of temperatures as in was in the first part (the most confusing phrase is 4:45 "as $A$ goes to $B$ is goes from being colder than $B$ to hotter than B", why? it just goes to the $B$). I don't understand how the assumption that there are a point on the track where the temperature is as in the point $B$ can help us, even if it's true is not what we need we need the temperature in the point $A$ to be equal the temperate in the point $B$.

The second assumption is to consider all antipodal points with the same temperature and consider all the points on the track with the same temperature of the opposite point, so as result we have a "club" of the intermediate point with the different temperatures, but all their temperatures equal to the temperature of the opposite point, given so how can we connect them by the line.

I have some problems in understanding the idea of the second part, would appreciate for help.

Answer 1

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.

Answer 2

Creator of the video here.

but along this part we don't move A and B there is no intersection of temperatures as in was in the first part

Yeah I didn't elaborate on this as much as the previous section, but the same thing is happening, just along an arbitrary path of connected opposite points, instead of a great sphere. What also might be unclear, is that B tracks the opposite point of A. Its path just isn't drawn.

I hope that helps. Vsauce also released a video very recently which runs through my explanation (I think he got it from me - I'm credited in his video description), so maybe it would also be useful: https://www.youtube.com/watch?v=csInNn6pfT4