TITLE: Euler Characteristic of a manifold with non-vanishing vector field,
QUESTION [26 upvotes]: A friend of mine recently asked me if I knew any simple, conceptual argument (even one that is perhaps only heuristic) to show that if a triangulated manifold has a non-vanishing vector field, then Euler's formula (the alternating sum of the number of faces of given dimensions) vanishes. I didn't see how to get started, but it seems like a good MO question.

REPLY [3 votes]: This is basically a watered down version of Poincare-Hopf theorem. Assuming that your compact manifold $M$ has a tringulation. At each of the simplex put a vector field $V$ which is zero at the centre of its subsimplices and flows out from the centre of the higher simplex to its boundary simplices. This vector field has finitely many zeroes and the index of $V$ is precisely $\chi(M)$. Now notice that the index of a vector field doesn't depend on the vector field. This is because it's topologically the intersection number of the image of $V$ in $TM$ and $M$ sitting as the zero section and $V$ is homotopic to $M$. In other words, given your non-vanishing vector field $V'$ one can linealy homotope this to $V$ whose index calculates $\chi(M)$.