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There is hint: if M has isolated singular points, find a diffeomorphism to make these singular points in a any neighborhood which you want. How can we do next?

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    @Javier Actually... you were right, it's just [degree theory](http://math.stackexchange.com/questions/47451/consequences-of-degree-theory), essentially.2011-06-26

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// Let me begin with an obstruction-theoretic solution — in hope that someone might find it of interest.

(For simplicity, let $\pi_1(M)=0$.) Non-vanishing vector field is a section of the spherization of $TM$, a bundle with fiber $S^{n-1}$. Obstructions to finding a section of a bundle with fiber $S^{n-1}$ lie in groups $H^k(M;\pi_{k-1}S^{n-1})$. These groups are trivial for $k (since coefficients are trivial) and for $k>n$ (since $M$ is $n$-dimensional). So the only non-trivial obstruction is the principal obstruction $\chi\in H^n(M;\pi_{n-1}(S^{n-1}))=\mathbb Z$.

And it's not hard to show, that it coincides with Euler char (indeed, the value of the obstruction on an $n$-cell is the degree of vector field on the bounding sphere — which coincides with the sum of indices of singular points of an extension of the field inside the cell).

// Reference (obstruction-theoretic approach to char. classes): Milnor-Stasheff, section 12.


This also explains, how to solve the problem directly (actually, it's the same solution in slightly different language). Take any vector field $v$ on $M$, and choose some sphere, containing all singular points. Degree (aka index) of the field on the sphere is exactly $\chi(M)=0$ — which means exactly that there is a non-vanishing extension of the field from the sphere to the ball (coinciding with $v$ on the boundary, but not inside the ball).

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    I don't see how having $\chi(M)=0$ means the extension is non-vanishing.2012-07-27