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Let $M^n$ be a smooth, compact manifold. Show that if $f:M\to\mathbb{R}^n$ is smooth, then $f$ is not a submersion.

Let $n=1$, $M=(0,1)$ and $f:x\mapsto x$, then $f_{*_{x}}=1\neq 0$ for all $x\in(0,1)$, so $f$ is a submersion. Isn't this a counter example?

Am I missing something? Thanks!

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    $\begingroup$ Is $(0,1)$ compact? ;) $\endgroup$ – Renan Maneli Mezabarba Nov 1 '16 at 20:07
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    $\begingroup$ thanks, @RenanManeliMezabarba, you're totally right. I'll rewrite the question $\endgroup$ – rmdmc89 Nov 1 '16 at 20:12
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    $\begingroup$ In addition to the answers you got, there is a more elementary proof: let $f^1$ be the value of the first coordinate of the map $f$. It is a smooth function on a compact manifold so it attains a supremum somewhere. At that point in particular you have that the Jacobian matrix $Df$ has vanishing first row, and cannot be full rank. $\endgroup$ – Willie Wong Nov 1 '16 at 20:58
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Let $f:M^n\rightarrow\mathbb{R}^n$ submersion and $M^n$ compact.

$\mathbb{R}^n$ is connected implies that $f$ is onto, therefore $f(M^n)=\mathbb{R}^n$.

On the other hand, as $f$ is continuous and $M^n$ compact then $f(M^n)$ is compact but $\mathbb{R}^n$ is not compact.

Contradiction, there is not $f$ a submersion.

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  • $\begingroup$ why does the connectedness of $\mathbb{R}^n$ imply that $f$ is onto? $\endgroup$ – rmdmc89 Nov 1 '16 at 20:27
  • $\begingroup$ The only open and closed at a time in a set connected are empty and the total. In this case $f(M^n)$ is closed and open simultaneously. $\endgroup$ – jimbo Nov 1 '16 at 20:32
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    $\begingroup$ @AguirreK" jimbo is using the fact that submersions are open maps (a fact with which you seemed to not be familiar). $\endgroup$ – Willie Wong Nov 1 '16 at 20:56
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As pointed out in the comments, $(0,1)$ is not compact, thus your function is not a counterexample.

Let me just add a sketch of the proof of the tittle of your question.

Recall that submersions are open mappings, while the compactness of $M$ implies that $f$ is closed. Thus, $f(M)$ is a nontrivial clopen of $\mathbb{R}^n$ , hence $f(M)=\mathbb{R}^n$ (the only nontrivial clopens of connected spaces are $\emptyset$ and the whole space). On the other hand, the compactness of $M$ also implies that $f(M)$ is compact, a contradiction.

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  • $\begingroup$ @AguirreK We have that $\emptyset\ne f(M)$ is a clopen, hence the only possibility is $f(M)=\mathbb{R}^n$. In fact, the same reasoning works if you replace $\mathbb{R}^n$ by a connected smooth manifold $N$. $\endgroup$ – Renan Maneli Mezabarba Nov 1 '16 at 20:41
  • $\begingroup$ Thanks, but it's still not clear to me why every submersion is an open map $\endgroup$ – rmdmc89 Nov 1 '16 at 20:42
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    $\begingroup$ Roughly speaking, a submersion is like a "local projection", and projections are open. $\endgroup$ – Renan Maneli Mezabarba Nov 1 '16 at 20:48
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    $\begingroup$ I guess yes. You may find this interesting. $\endgroup$ – Renan Maneli Mezabarba Nov 1 '16 at 20:58
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    $\begingroup$ @AguirreK Compact subspaces of Hausdorff spaces are closed. Does it help? $\endgroup$ – Renan Maneli Mezabarba Nov 1 '16 at 21:28

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