Why is an antipodal-symmetrically colored circle guaranteed to have an odd number of multicolored edges? I'm reading a proof of 2D Tucker's Lemma.  It asserts the following claim without proof:
Drop points on a circle in antipodal fashion (i.e. if there is a point at position $p$, then there also must be a point on the other end of the diameter through $p$).  Label each point on the circle with one element of the set $\{-1, 1, -2, 2\}$ such that if a point is labeled $x$, then its antipodal point is labeled $-x$.  Conclusion: If no arcs have endpoints labeled $(-1, 1)$ or $(-2, 2)$ (in either order), then there are an odd number of arcs labeled $(-1, 2)$.
I can't figure out why this is true.  Any ideas?
 A: I'm pretty sure what you have written in the body of the question is not correct.  For example this arrangement, read going around the circle, should be acceptable: $1,-2 ,1 , -1,2 ,-1$.  This has two edges 'with endpoints labelled -1 and 2 (in either order)'.  Am I misunderstanding something?
A: If two vertices labelled $1$ are adjacent, collapse them to a single vertex labelled $1$, and collapse the antipodal vertices to a single vertex labelled $-1$. Repeat until there are no adjacent vertices labelled $1$. Then do the same thing with adjacent vertices labelled $2$. You still have a properly labelled division of the circle, and all edges now have labels $\langle -1,2\rangle$, $\langle 1,-2\rangle$, $\langle 1,2\rangle$, or $\langle -1,-2\rangle$ in one order or the other. I’ll refer to an edge whose endpoints are $-1$ and $2$ (in either order) as an edge of type $\langle -1,2\rangle$, and similarly for the other three types. As you traverse the circle, the absolute values of the labels now alternate between $1$ and $2$.
Suppose that there is no edge of type $\langle 1,2\rangle$; then every edge is a sign-changing edge. But every edge also changes the absolute value of the label, so either the labels of the vertices are alternately $1$ and $-2$, or they are alternately $-1$ and $2$, neither of which is possible.
Thus, there is an edge $e$ of type $\langle 1,2\rangle$; its antipodal edge $e'$ is a type $\langle -1,-2\rangle$ edge. Consider the arc of the circle from the $2$ vertex of $e$ to the $-1$ vertex of $e'$. This arc must have an odd number of edges, since the absolute values of the labels at its ends are different. Their signs are also different, so there must be an odd number of sign-changing edges, i.e., edges of types $\langle -1,2\rangle$ and $\langle 1,-2\rangle$. Suppose that there are $m$ edges of type $\langle -1,2\rangle$ and $n$ of type $\langle 1,-2\rangle$. Then the antipodal arc has $n$ of type $\langle -1,2\rangle$ and $m$ of type $\langle 1,-2\rangle$, and $m+n$ is odd, so there are an odd number of each of the types $\langle -1,2\rangle$ and $\langle 1,-2\rangle$.
