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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!

rmdmc89
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    Is $(0,1)$ compact? ;) – Renan Mezabarba Nov 01 '16 at 20:07
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    thanks, @RenanManeliMezabarba, you're totally right. I'll rewrite the question – rmdmc89 Nov 01 '16 at 20:12
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    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. – Willie Wong Nov 01 '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.

jimbo
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  • why does the connectedness of $\mathbb{R}^n$ imply that $f$ is onto? – rmdmc89 Nov 01 '16 at 20:27
  • 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. – jimbo Nov 01 '16 at 20:32
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    @AguirreK" jimbo is using the fact that submersions are open maps (a fact with which you seemed to not be familiar). – Willie Wong Nov 01 '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.

Renan Mezabarba
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  • @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 $N$. – Renan Mezabarba Nov 01 '16 at 20:41
  • Thanks, but it's still not clear to me why every submersion is an open map – rmdmc89 Nov 01 '16 at 20:42
  • Well, I think that by definition it is a different question. – Renan Mezabarba Nov 01 '16 at 20:44
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    Roughly speaking, a submersion is like a "local projection", and projections are open. – Renan Mezabarba Nov 01 '16 at 20:48
  • Oh, I see, locally open maps are globally open, right? That really helped, thanks! – rmdmc89 Nov 01 '16 at 20:52
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    I guess yes. You may find this interesting. – Renan Mezabarba Nov 01 '16 at 20:58
  • @RenanManeliMeyabarba, in case we substitute $\mathbb{R}^n$ by a connected smooth manifold $N$, we well have that $f(M)$ is open and compact. But can we guarantee that $f(M)$ is closed? I think compactness doesn't imply closure in general, right? – rmdmc89 Nov 01 '16 at 21:23
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    @AguirreK Compact subspaces of Hausdorff spaces are closed. Does it help? – Renan Mezabarba Nov 01 '16 at 21:28
  • OK, at the risk of being laughed at because this thread is so old, something is bothering me. Suppose $M^n$ is the closed unit ball of $R^n$ and $f$ is the identity. Then $f$ is a smooth submersion (right?) but $f(M^n)$ = $M^n$ \neq R^n$. The problem for me is that an open map doesn't HAVE to be strongly open, just relatively open. My feeling is that the posters here are right and I'm wrong, but I'd really appreciate someone telling me why? Thank you in advance. – user167131 Jan 21 '25 at 17:07
  • @user167131 I don’t recall any results related to submersions (at the time this question was raised, it was inside the scope of a course I was taking as a student). My suggestion: consider rephrasing your comment as a question. Apologies for not being more helpful. – Renan Mezabarba Jan 21 '25 at 17:55
  • No, that's very helpful, thank you. I just don't like to get "jumped on" for asking a dumb question (I guess I think Comments go under the radar). :) – user167131 Jan 21 '25 at 17:57