It is stated in may books and also in the wikipedia or WolframMathWorld that the Klein bottle is not homeomorphic to a subspace of $\mathbb R^3$. Why is it so?
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1Related: How does one prove that the Klein bottle cannot be embedded in $R^{3}$? – Andrew D. Hwang May 09 '17 at 11:53
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1Also this one on MathOverflow: https://mathoverflow.net/questions/18987/why-cant-the-klein-bottle-embed-in-mathbbr3 – Lee Mosher May 09 '17 at 11:57
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If so, one could apply Alexander Duality to conclude that $$H_2(\text{Klein bottle},\mathbb{Z}) \approx \widetilde H_0(\mathbb{R}^3 - (\text{Klein bottle})) \approx \mathbb{Z} $$ contradicting that $H_2(\text{Klein bottle},\mathbb{Z})$ is trivial.
Lee Mosher
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