Is there a simple topological proof for a topological theorem about $S^2$? Consider the problem of coloring each point of $S^2$ with one of two colors (say "black" or "white") so that among any three points of $S^2$ which are the vertices of an equilateral spherical triangle with 90 degree sides, one and only one of these three points will be colored "white". This problem comes from Quantum Mechanics and a theorem of Quantum Mechanics, the Kochen-Specker theorem, has-as one of its consequences-that no such coloring of the points of $S^2$ is possible. But all the proofs of this result that I have seen are based on constructing specific finite subsets of $S^2$ (the smallest containing 33 points with specified real x-y-z coordinates) and showing that the coloring problem cannot be solved for these finite sets. The points apply combinatorial analysis to finite graphs whose vertices are the points of these sets. Two points which are 90 degrees apart on $S^2$ are joined by an edge in the graphs. The hard part of the proofs is finding these finite sets in the first place. Also, the combinatorial work is unwieldy to specify and carry out in detail........But the coloring theorem is really a (rather interesting) theorem about the metric topology of $S^2$. It seems to me that it should have a topological (or at least a coordinate-free) proof. Such a proof might even be alot simpler to follow than the proofs which involve specific coordinates. Does such a proof exist?
 A: If I understand the problem correctly, one wants a coloring of the points of the unit sphere $S^2$ so that for every triangle with edge lengths $\pi/2$, precisely two vertices are colored black and one is colored white? 
If that's correct, then I think there might be a topological proof. If one takes a black point $p$ , then there is a great circle $p^{\perp}$ which is distance $\pi/2$ away. For every point on $p^{\perp}$ colored white, the two points on $p^\perp$ distance $\pi/2$ away must be colored black, and vice versa (in fact, antipodal points must have the same color). Also, for every white point $q$, the circle $q^\perp$ must be colored entirely black. 
From this, one sees that for every black point $r$ on $p^\perp$, the great circle going through $r$ and $p$ must be entirely black, since it is $q^\perp$ for some point $q\in p^\perp$ distance $\pi/2$ from $r$. Consider another black point $s\in p^\perp$ distance $<\pi/2$ from $r$, and distance $\pi/2$ to $t\in p^\perp$ colored white. The $t^\perp$ is also black. For every point $u\in q^\perp$, there are a pair of points in $t^\perp$ which are distance $\pi/2$ from $u$. There is a unique antipodal pair of white points distance $\pi/2$ from these, and together these form a curve of white points going through $q$. The great circles through $p$ which are colored black cannot cross this white curve, and therefore there is an interval about $q$ in $p^\perp$ which is colored white. However, this implies that the set of white points in $p^\perp$ is an open set. Therefore, so is the set of black points. But $p^\perp$ is connected, so it cannot be a union of two open subsets, a contradiction. 
