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 $\mathbb{R}^k$?
Best Answer
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.