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?
Image of smooth manifold is a submanifold
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analysis
differential-topology
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2Take an open interval and map it into the plane such that it forms a figure "6". Focus on that non-manifold point on the 6 and think about how you can make something like that and still have an injective mapping. – 2012-10-04
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0You can't do that with a proper map http://en.wikipedia.org/wiki/Proper_map – 2012-10-04