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Can anyone give me an hint:

If $M$ is a compact manifold of dimension $n$ and $f:M\rightarrow \mathbb{R}^n$ is $C^{\infty}$, then $f$ can not be everywhere nonsingular.

Suppose $f$ is everywhere non-singular. Then $df_m:T_m(M)\rightarrow \mathbb{R}^n_{f(m)}$ would be an isomorphism, right? But I do not understand where is the contradiction.

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$\mathbb{R}^n$ is a manifold of dimension $n$ and the identity map is smooth. –  Neal Jun 28 '12 at 21:37
Mex: I edited the ${\mathbb C}^\infty$ to $C^\infty$; the notation ${\mathbb C}$ means complex numbers. –  KCd Jun 28 '12 at 21:47
Consider the canonical coordinate functions on $\mathbb{R}^n$, pulled back to $M$. $M$ is compact, so these functions must attain their maxima and minima. Therefore... –  Zhen Lin Jun 28 '12 at 22:24
Just project onto the first coordinate. You then get a smooth function $M\to \mathbb{R}$. It attains its max, so now conclude something about the differential. –  Matt Jun 28 '12 at 23:23
Alternately, if $f$ has no singular points, then the image of $M$ is open (by the inverse function theorem). But by compactness it is closed... –  Qiaochu Yuan Jun 29 '12 at 1:58

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