Let $ABC$ be a triangle, and let $S$ be the set of vertices of some triangulation $T$ of $ABC$ . Let $f$ be a mapping of $S$ into a three-element set, say $\{1,2,3\}=T$ (indicated by red/green/blue respectively in the figure), such that:

• any point $P$ of $S$ , if it is on the side $AB$ , satisfies $f(P)\in\{1,2\}$
• similarly if $P$ is on the side $BC$ , then $f(P)\in\{2,3\}$
• if $P$ is on the side $CA$ , then $f(P)\in\{3,1\}$

We will informally sketch a proof of a stronger statement: Let $M$ (resp. $N$ ) be the number of simplexes satisfying (1) and whose vertices have the same orientation as $ABC$ (resp. the opposite orientation). Then $M-N=1$ (whence $M>0$ ).

The proof is in the style of well-known proofs of, for example, Stokes's theorem in the plane, or Cauchy's theorems about a holomorphic function.

Define an antisymmetric function $d:T\times T\to\mathbb{Z}$ by $$d(1,1)=d(2,2)=d(3,3)=0$$ $$d(1,2)=d(2,3)=d(3,1)=1$$ $$d(2,1)=d(3,2)=d(1,3)=-1\;.$$

Let's define a ``circuit'' of size $n$ as an injective mapping $z$ of the cyclic group $\Z/n\Z$ into $V$ such that $z(n)$ is adjacent to $z(n+1)$ for all $n$ in the group.

Any circuit $z$ has what we will call a contour integral $Iz$ , namely $$Iz=\sum_n d(z(n),z(n+1))\;.$$ Let us say that two vertices $P$ and $Q$ are equivalent if $f(P)=f(Q)$ .

There are four steps:

1) Contour integrals are added when their corresponding circuits are juxtaposed.

2) A circuit of size 3, hitting the vertices of a simplex $PQR$ , has contour integral

• 0 if any two of $P$ , $Q$ , $R$ are equivalent, else
• +3 if they are inequivalent and have the same orientation as $ABC$ , else
• -3

3) If $y$ is a circuit which travels around the perimeter of the whole triangle $ABC$ , and with same orientation as $ABC$ , then $Iy=3$ .

4) Combining the above results, we get $$3=\sum Iw=3M - 3N$$ where the sum contains one summand for each simplex $PQR$ .

Remarks: In the figure, $M=2$ and $N=1$ : there are two ``red-green-blue'' simplexes and one blue-green-red.

With the same hypotheses as in Sperner's lemma, there is such a simplex $UVW$ which is connected (along edges of the triangulation) to the side $AB$ (resp. $BC$ ,$CA$ ) by a set of vertices $v$ for which $f(v)\in\{1,2\}$ (resp. $\{2,3\}$ , $\{3,1\})$ . The figure illustrates that result: one of the red-green-blue simplexes is connected to the red-green side by a red-green ``curve'', and to the other two sides likewise.

The original use of Sperner's lemma was in a proof of Brouwer's fixed point theorem in two dimensions.