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I'm working with the definition:

A smooth embedding of M into N is an injective immersion $F:M\rightarrow N$ that is also a topological embedding.

I'd like to explain why the klein bottle cant embed in $\mathbb{R}^3$. The immersion is not injective, thus results in self intersections, correct?

Jonathan
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  • If you have self-intersection in a point $P$, how do you cope with the non-unicity of tangent space in $P$, for example ? – Jean Marie Feb 02 '17 at 18:18
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    This reasoning raises the question why the immersion is not (cannot be) injective. – Thomas Feb 02 '17 at 18:25
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    Are you asking why the "standard picture" of the Klein bottle is not an embedding? Or, instead, how to prove that there does not exist any embedding? – Lee Mosher Feb 02 '17 at 19:13
  • It suffices to make it kleiner und kleiner, thus at the end, we will not see the self intersection... – Jean Marie Feb 02 '17 at 22:05
  • Argue that $F$ cannot be an immersion since that Klein bottle contains a copy of the Mobius strip. – Faraad Armwood Feb 05 '17 at 02:30

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