Hairy Ball theorem and its applications While searching a question about fibre bundles, which was asked here, i got directed to Vector bundles. I noticed this word "Hairy Ball" which sounded eccentric and made a search at Wikipedia. 
How is the hairy ball theorem related to this statement: You can't comb the hair on a coconut.
 A: More of a comment than an answer, but it didn't fit:
In fact, the hairy ball theorem works on any even-dimensional sphere.  However, odd-dimensional spheres do admit nonvanishing vector fields: if you use the usual embedding of $S^{2k-1}$ in $\mathbb{R}^{2k}$, then at the point $(x_1,y_1,\ldots,x_k,y_k)$ you can put the (nonzero) tangent vector $(y_1,-x_1,\ldots,y_k,-x_k)$.
The natural generalization is: How many linearly independent tangent vector fields can exist on $S^{n-1}$?  As it turns out, the answer only depends on how many factors of 2 there are in $n$.  So e.g. $S^3$ and $S^{35}$ both admit exactly 3 linearly independent tangent vector fields, because $3+1=2^2$ and $35+1=2^2\cdot 9$.  This is crazy!
A: I think the picture at the top right of the wikipedia page does a pretty good job at visualization of this.
The idea is that if you think of each hair on a coconut as a vector with base at a point on the sphere, then we have a notion of what it means for these to change continuously. So combing the hairs is essentially the analogy for making all the hairs lie tangent to the sphere and for them to vary continuously. This cannot be done, since if you have a continuous vector field on the sphere, it must be zero somewhere. The way the picture shows this is that you get hairs that point outward or "normal" to the sphere (which I think is not a really good way to think about the theorem) and hence have not been combed. 
Notice that you can't just comb that one part down, or else it won't be continuous. There will be some "overlapping".
