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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.2012-06-28
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    Mex: I edited the ${\mathbb C}^\infty$ to $C^\infty$; the notation ${\mathbb C}$ means complex numbers.2012-06-28
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    sorry,M is compact, Kcd: I will never do this mistake again.2012-06-28
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    Makes more sense now! I tried to make the title more descriptive.2012-06-28
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    Could you please recall the definition of a singular point of a map.2012-06-28
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    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...2012-06-28
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    what is pull back of coordinate? let assume $(x_1,\dots,x_n)$ be the coordinate function on $\mathbb{R}^n$, then?2012-06-28
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    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.2012-06-28
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    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...2012-06-29

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