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Conisder a straight simplex $\Delta^n$ in $\mathbb R^n$ and take a generic constant vector field $v$ (transversal to the faces of $\Delta^n$). Choose all faces of $\Delta^n$ such that the field moves the center of the face inside the simplex. Then the alternating sum of the numbers of these simplexes (signed by the parity of the dimension) is zero(this is true because $(1-1)^n=0$)..

Added

Edited. Let me explian the meaning There was an explanation here with a mistake (spoted by Sergei) of $(1-1)^n$. If we look on how this constant generic vector field $v$ intersects $\Delta^n$, it why each simplex contributes zero, but the statement is not hard to see that there will be correct. The new proof is a unique vertex $A$ of follows: $\Delta^n$ such (-1)^{n-1}+(-1)^n=0$.

Proof. Let us say that the trajectory of $v$ (is the sunlight. Then it enlightens a line) through $A$ intersects part of the opposite to $A$ face $\Delta^{n-1}$. Now we have two possibilites:

1) simplex $A$ is pushed away \Delta^{n+1}$. Consider the shade from $\Delta^n$ by $v$, in this case only $\Delta^{n-1}$ and $\Delta^{n}$ contribute to on some plane below the sum, and we get $1-1=0$.

2) $A$ simeplx. The shade is moved by $v$ inside $\Delta^n$, then an convex set. It is naturally decomposed into simplexe, so the faces sum of simplexes over this shade is $\Delta^n$ that will contribute to (-1)^{n-1}$ (because the sum are exactly simplexes in the faces that are adjucent to $A$ -- one vertex, $n$ edges, $\frac{n(n-1)}{2}$ two faces, ectboundary of this convex set do not contrubute). So And we also get the binomial sum $(1-1)^n$.(-1)^n$ for $\Delta^n$.

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Conisder a straight simplex $\Delta^n$ in $\mathbb R^n$ and take a generic constant vector field $v$ (transversal to the faces of $\Delta^n$). Choose all faces of $\Delta^n$ such that the field moves the center of the face inside the simplex. Then the alternating sum of the numbers of these simplexes (signed by the parity of the dimension) is zero (this is true because $(1-1)^n=0$).

Now, if you have a fine enough triangulation of $M$ and a vector field transversal to all faces, we can apply the above reasoning to the whole manifold.

Added. Let me explian the meaning of $(1-1)^n$. If we look on how this constant generic vector field $v$ intersects $\Delta^n$, it is not hard to see that there will be a unique vertex $A$ of $\Delta^n$ such that the trajectory of $v$ (a line) through $A$ intersects the opposite to $A$ face $\Delta^{n-1}$. Now we have two possibilites:

1) $A$ is pushed away from $\Delta^n$ by $v$, in this case only $\Delta^{n-1}$ and $\Delta^{n}$ contribute to the sum, and we get $1-1=0$.

2) $A$ is moved by $v$ inside $\Delta^n$, then the faces of $\Delta^n$ that will contribute to the sum are exactly the faces that are adjucent to $A$ -- one vertex, $n$ edges, $\frac{n(n-1)}{2}$ two faces, ect. So we get the binomial sum $(1-1)^n$.
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Conisder a straight simplex $\Delta^n$ in $\mathbb R^n$ and take a generic constant vector field $v$ (transversal to the faces of $\Delta^n$). Chose Choose all faces of $\Delta^n$ such that the field moves the center of the face inside the simplex. Then the alternative alternating sum of the numbers of these simplexes (signed by the parity of the dimesniondimension) is zero (this is true because $(1-1)^n=0$).

Now, if you have a fine enought enough triangulation of $M$ and a vector field transveral transversal to all faces, we can apply the above reasoning to the whole manifold.

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