Divide the 1-dimensional sphere $S^1$ into arcs. 
Label each vertex between two arcs as either 1 or 2. 
Count the number of "full arcs" - arcs with both 1 and 2 at their endpoints. 
Divide these full arcs to two kinds: "positive" (1-2 going clockwise) and "negative" (2-1).
Then, it is clear that the number of positive full arcs equals the number of negative full arcs: each time we go from 1 to 2, we must at some point go back from 2 to 1, since we are on a circle.



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> Is the following generalization to multi-dimensional spheres correct?
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Triangulate the $n$-dimensional sphere $S^n$.
Label each vertex of the triangulation by a label from $\{1,\ldots,n+1\}$.
Count the number of "full simplices" - simplices with all $n+1$ different labels at their vertices.
Divide labeled simplices into two equivalence classes: "positive" and "negative", such that for each two simplices in the same class, there is an orientation-preserving map from one to the other agreeing on the labels. To relabel a simplex keeping it in the same class, we can perform any *even permutation* on the labels (for example, if a simplex labeled 1-2-3-4 is positive, then the simplex labeled 2-3-1-4 is also positive, but the simplex labeled 2-1-3-4 is negative). 



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> Conjecture: the number of positive simplices equals the number of negative simplices.
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Is this conjecture correct?


Note: [asked in math.SE with no replies](https://math.stackexchange.com/q/2043284/29780)



Labeling a triangulated sphere
The labeling gives a piecewise linear map from $S^n$ to a simplex $\Delta$ with $n+1$ vertices, where label $k$ means the point is sent to vertex $v\_k$, and we extend linearly on simplices. The map must have degree $0$ because $\Delta$ is contractible. The degree can be computed from the preimage of any point in the interior of $\Delta$, and it is the difference between the number of positive simplices and the number of negative simplices. 


