Every manifold admits a vector field with only finitely many zeros Let $M$ be a smooth manifold. I am trying to prove that $M$ admits a vector field with only finitely many zeros. 
This will follow if we can find a function $f : M\rightarrow \mathbb R$ such that $df$ has only finitely many zeros, but I cannot find such a function with this property either. My initial idea was to try to embed $M$ in $\mathbb R^N$ for some $N$ and look at $x\mapsto u \cdot x$ for fixed $u\in \mathbb R^N$, but I could not find a way to prove that there must be a $u$ such that the differential of this map has only finitely many zeros. 
Does anyone have an elementary construction of such a vector field (or function)?
 A: A classic method - I think of Steenrod -  is to triangulate the manifold then form the vector field whose singularities are the barycenters of the triangulation.For instance on a triangle the field flows away from the barycenter of the triangle towards the vertices and the centers of the edges. along the edges the field flows away from the centers towards the vertices. Draw a picture. It is easy to see.
A: Here's a fun proof that employs the Transversality theorem to show that on any smooth manifold $M$, there is a vector field that vanishes only on a $0$-dimensional submanifold of $M$. Of course, when $M$ is compact, every $0$-dimensional submanifold is finite, which gives you your desired result.
Assume without loss of generality that $M^n$ is embedded in $\mathbb{R}^N$ with $N>n$. Define a map $F:M\times \mathbb{R}^N\to TM$ by $F(p,v)=\text{proj}_{T_pM}v$. Then $F$ is a smooth submersion. In particular, $F$ is transverse to $Z=M\times \{0\}$. So, by the transversality theorem, there exists some $v\in \mathbb{R}^N$ so that $f_v(x):M\to TM$ is transverse to $Z$. Now, $f_v(x)$ is a smooth section of $TM$, and so $f_v$ is a vector field. So $f_v^{-1}(M\times \{0\})$-the zeros of $f_v$-is a submanifold of $M$ of codimension $\dim TM-\dim M\times \{0\}= \dim M$, as claimed.
