Let $M$ be a smooth, n-dimensional manifold. Prove that for every $k \leq n$ there exists an embedding $ \mathbb{R}^k \to M$.

I'm having trouble visualising this. How can $\mathbb{R}^2$ be embedded into the 2-dimensional torus, considering that the torus is a compact manifold and $\mathbb{R}^2$ is certainly not. $f(\mathbb{R}^2)$ is supposed to be a submanifold of the torus and there are not that many options left. Is there something obvious I'm forgetting?

  • $\begingroup$ Your unease with $\mathbb{R}^2$ and the torus is warranted. If $f: X\to Y$ is an embedding and $Y$ is compact, then the properness of $f$ implies that $f^{-1}(Y)$, i.e. $X$ itself, is compact. So the statement cannot be true in general (if you assume embeddings are proper). Where did you find the statement/problem? $\endgroup$ – Kyle Jul 3 '15 at 4:47
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    $\begingroup$ Embeddings are proper iff they are closed, which need not be the case in general. $\endgroup$ – archipelago Jul 5 '15 at 12:37
  • $\begingroup$ A manifold has embedded charts by definition, and into these we may embed Euclidean spaces of smaller dimension. $\endgroup$ – Urs Schreiber May 12 '17 at 15:04

An open disc on any surface is always diffeomorphic to the whole $\Bbb{R}^2$. To see that, take for example the smooth, invertible map (in polar coordinates): $$ \theta \mapsto \theta\\ r\mapsto \arctan r. $$

  • $\begingroup$ But I need a diffeomorphism onto its image, don't I? $\endgroup$ – Cosmare Jul 1 '15 at 14:48
  • $\begingroup$ Sure, sorry, edited. $\endgroup$ – geodude Jul 1 '15 at 14:48
  • $\begingroup$ So then I can take $\Bbb{R}^k$ and it's diffeomorphic to an open disc on the manifold and that finishes the proof? $\endgroup$ – Cosmare Jul 1 '15 at 14:51
  • $\begingroup$ Yeah, I think so. $\endgroup$ – geodude Jul 1 '15 at 14:55

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