Sperner’s lemma
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 threeelement set, say $\{1,2,3\}=T$ (indicated by red/green/blue respectively in the figure), such that:

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

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similarly if $P$ is on the side $BC$, then $f(P)\in \{2,3\}$

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if $P$ is on the side $CA$, then $f(P)\in \{3,1\}$
(It follows that $f(A)=1,f(B)=2,f(C)=3$.) Then some (triangular) simplex of $T$, say $UVW$, satisfies
$$f(U)=1\mathit{\hspace{1em}\hspace{1em}}f(V)=2\mathit{\hspace{1em}\hspace{1em}}f(W)=3.$$ 
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 $MN=1$ (whence $M>0$).
The proof is in the style of wellknown 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 $\mathbb{Z}/n\mathbb{Z}$ 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

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0 if any two of $P$, $Q$, $R$ are equivalent, else

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+3 if they are inequivalent and have the same orientation as $ABC$, else

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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
$$3=\sum Iw=3M3N$$ 
where the sum contains one summand for each simplex $PQR$.
Remarks: In the figure, $M=2$ and $N=1$: there are two “redgreenblue” simplexes and one bluegreenred.
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 redgreenblue simplexes is connected to the redgreen side by a redgreen “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^{}.
Title  Sperner’s lemma 

Canonical name  SpernersLemma 
Date of creation  20130322 13:44:33 
Last modified on  20130322 13:44:33 
Owner  mathcam (2727) 
Last modified by  mathcam (2727) 
Numerical id  12 
Author  mathcam (2727) 
Entry type  Theorem 
Classification  msc 55M20 