TITLE: A smooth function $f:S^1\times S^1\to \mathbb R$ must have more than two critical points.
QUESTION [8 upvotes]: I am trying to show that a smooth function $f:S^1\times S^1\to \mathbb  R$ must have more than two critical points. Since $f$ attains maximum and minimum, it must have at least two critical points. How would one show that they can't be two? 
If one considers the gradient vector field $\nabla f$ then by the Poincare Hopf index theorem its index is equal to the Euler characteristic of the torus, i.e. it is $0$. Therefore $\nabla f$ has an even number of zeros. 
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REPLY [6 votes]: The only closed manifolds which allow a function with two (maybe degenerate) critical points are spheres. In dimension 2 it is quite easy to prove.
One of these critical points is maximum and another is minimum. Any level set in between is compact and has dimension 1, so it is disjoint union of circles. Now use an argument from Morse theory: because there is no critical points between two such level sets, the gradient vector field gives diffeomorphism of them. So the whole space is the suspension over disjoint union of circles. It is not difficult to show that it is manifold only in case of one circle, and then it is 2-sphere, not torus.