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The following should be something well?-known, but i haven't seen it anywhere, neither have i met any references about.

Let $M^{n}$ be a $n$-dimensional oriented closed manifold with a (sufficiently small) triangulation $\tau$. We "colour" the vertices of $\tau$ with $n+2$ colors: $v^{o}\rightarrow w(v^{o})\in$ {$1,2,...,n+2$ } and we shall say that the correspondence $w$ is a "coloring" of $\tau$. Take an arbitrary color $i\in$ {$1,2,...,n+2$ } and consider the $n$-simplices whose vertices are colored with exactly the colors {$1,2,...,n+2$ }$\backslash\{i\}$. Let $\Delta^{n}$ be such a simplex and $v_{1},...,v_{n+1}$ be its vertices ordered according to the positive orientation of $\Delta^{n}$ induced by the orientation of $M^{n}$. Then we write $\sigma_{i}(\Delta^{n})=1$, if the permutation $(w(v_{1}),...,w(v_{n+1}))$ is even, and $\sigma_{i}(\Delta ^{n})=-1$ otherwise. Set $\sigma_{i}(\Delta^{n})=0$ if some vertex of $\Delta^{n}$ is colored $i$, or there are two identically colored vertices. Let finally


where the sum is over all $n$-simplices.

The Claim: The number $\sigma_{i}(w)$ does not depend on $i$: $\sigma_{1}(w)=\sigma_{2}(w)=...=\sigma_{n+2}(w)$. So we have a global invariant $\sigma(w)$ of the coloring $w$.

This invariant has a geometrical meaning: Consider the dual cell complex of the triangulation $\tau$, then since each cell corresponds to a vertex $v^{o}$ of $\tau$, we may color this cell by the color $w(v^{o})$. Let $F_{i}$ be the union of all cells colored $i$, then we get a covering $\lambda=\{F_{1},...,F_{n+2}\}$ of $M^{n}$. It is easy to see that the intersection of all $F_{i}$ is empty, so the canonical map of $M^{n}$ into the nerve of $\lambda$ may be considered as a map of $M^{n}$ into the $n$-sphere $\mathbb{S}^{n}$: $\varphi:M^{n}\rightarrow\mathbb{S}^{n}$. Then the degree of $\varphi$ equals $\sigma(w)$:


As the proofs are not sophisticated at all and the construction seems conceptual, maybe it is worth including this material in an elementary topology textbook. Note also that it gives a method for calculating the degree without smooth approximation.

Of course, i don't want to repeat well-known things without citation, so any references are welcome.

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there is a much simpler way to say the geometrical meaning: consider the standard $n+1$-simplex, with vertices $1, \ldots, n+2$. Then a coloring of the vertices of a triangulation is just the same thing as a simplicial map to this simplex. now the statement about the degree is obvious. – Vivek Shende Jan 27 '13 at 7:17

This is closely related to the Generalized Sperner's Lemma, which holds for all for simplicial manifolds with or without boundary. See my old survey for a quick introduction (Section 8.1). Classical references include A.B. Brown and S.S. Cairns, Strengthening of Sperner's lemma applied to homology theory, PNAS, 1960, and D.I.A Cohen, On the Sperner lemma, JCT (1967). I don't immediately see how your result follows from the lemma, but recall that many extensions and generalizations are known. I would start with these references and search forward to find your particular version.

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