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I'm having a little trouble seeing how to do Exercise 7.5 in Lee Smooth Manifolds:

Let $M$ be a smooth compact manifold. Show there is no submersion $F:M\rightarrow\mathbb{R}^k$ for any $k>0$.

If $F:M\rightarrow\mathbb{R}^k$ were a submersion, then $\dim(M)\geq k$. This rules out things like $\mathbb{S}^{k-1}\hookrightarrow\mathbb{R}^k$. Approaching it the other way, $\mathbb{B}^k\hookrightarrow\mathbb{R}^k$ is a submersion, but the open ball $\mathbb{B}^k$ is not compact. It seems like what's going on is that, since the image of $F$ would be a compact, hence closed, subset of $\mathbb{R}^k$, if $F(M)$ were "$k$-dimensional" it would require $M$ to have been a manifold with boundary, which isn't allowed. However, I'm not sure how to fill in the gaps here / make it rigorous.

On a possibly related note: Is it possible to have an immersed compact $k$-submanifold of $\mathbb{R}^k$?

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A codimension 0 immersion is a submersion, and so for closed manifolds (I assume you want compact without boundary) is equivalent to a covering map. (Use the rank-nullity theorem...) So what you just proved about submersions also apply. –  Willie Wong May 15 '11 at 23:13
    
@Willie: Yeah, I just realized that :) Well, two birds with one stone... –  Zev Chonoles May 15 '11 at 23:16
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up vote 4 down vote accepted

Submersions are open maps; but the image of $M$ is compact in a Hausdorff space, and hence closed as well. So it's a clopen nonempty set. Since $\mathbf{R}^n$ is connected, it's the whole thing. But then $\mathbf{R}^n$ is the quotient of a compact space, so it's compact, which is not true.

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What's your opinion of Lee's book, btw? –  Robert Haraway May 15 '11 at 23:03
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I really like it so far. I feel like it's kind of a Dummit and Foote, for differential topology. –  Zev Chonoles May 15 '11 at 23:09
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