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It's know that if $M$ is a compact, smooth manifold of dimension $n$ and the map $f: M \to \mathbb{R^m}$ is injective, smooth, $n \le m$ and $Jf(a)$, the Jacobian, has rank $n$ for every $a \in M$, then $f(M)$ is a submanifold of $\mathbb{R^m}$. I'm trying to think of a counterexample to this statement if we suppose $M$ is not compact, but haven't gotten anywhere. Can anyone offer a hint?

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    You can't do that with a proper map http://en.wikipedia.org/wiki/Proper_map2012-10-04

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Let $g\colon \mathbb{R} \to \mathbb{T}^2= \mathbb{R}^2/\mathbb{Z}^2$ be given by $t\mapsto [(t,\alpha t)]$, where $\alpha$ is any irrational number. The image of $g$ is a dense subset of $\mathbb{T}^2$ but $g$ is not surjective. Hence, $\mathrm{Im}(g)$ cannot be a submanifold of $\mathbb{T}^2$. Compose this map with the embedding of the torus into $\mathbb{R}^3$.