This question is for an homework. It follows like this :

Let $M$ be an $n$-dimensional manifold. Construct a smooth surjective map from $M$ to the torus $(S^1)^n$.

My first idea was to use the smooth covering map $ \epsilon^n : \mathbb{R}^n \rightarrow (S^1)^n : (x_1 , ..., x_n) \mapsto (e^{2\pi i x_1}, ... , e^{2\pi i x_n})$. In fact, with this and a chart $(U, \phi)$ of $M$ (and then $\phi (U) \cong \mathbb{R}^n$), I wanted $ \epsilon^n \circ \phi$ to be the surjective map I'm looking for. But it doesn't work because it is not defined on $M$ and I dont know how to fix this.

Maybe the solution doesn't need the use of the smooth covering map $\epsilon^n$ but I'm convinced that it's a good start.

Does someone have any ideas to help me? Thank you.

  • 2
    $\begingroup$ You can just multiply a bump function in front of your $\epsilon^n$. $\endgroup$ – cjackal Aug 30 '17 at 4:33
  • $\begingroup$ Can you be more specific? I don't understand your suggestion. $\endgroup$ – Sov Aug 30 '17 at 12:14
  • 1
    $\begingroup$ Ah, I misunderstood your notation. Anyway, an idea is essentially the same as yours. it is clear that you need to find a map $M\to \mathbb{R}^n$ whose image covers a cube of sidelength 1 in $\mathbb{R}^n$. As you noticed, you can use a chart, but the problem is this is not well-defined away from the interior of $\phi(U)$. So what you should do is multiplying a bump function, say $\rho$, supported on the interior of $\phi(U)$ to $\phi$ (coordinatewise of course) so that it is well-defined globally on $M$ and still has the covering property as you wish. $\endgroup$ – cjackal Aug 30 '17 at 17:33
  • $\begingroup$ It should have been $U$, not $\phi(U)$, in the above comment. $\endgroup$ – cjackal Aug 30 '17 at 17:54
  • $\begingroup$ Ok, now I got it. I was just confused with the word "multiply", it is more a composition of functions than a multiplication. Happy to see that my intuition was good, thanks for your help ! $\endgroup$ – Sov Aug 30 '17 at 18:12

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