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Let $M$ be the Moebius vector bundle over $S^1$.

Is it possible to embedd the total space of $M\oplus (S^1\times \mathbb{R^1})$ over $S^1$ into $\mathbb{R}^3$?

I suppose this isn't possible but I don't know an argument.

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up vote 3 down vote accepted

It's not possible, since the Whitney sum would be a nonorientable $3$-manifold, which cannot be embedded in $\mathbb R^3$.

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What's the reason that a nonorientable $3$-manifold cannot be embedded in $\mathbb{R}^3$? – André Feb 13 '12 at 22:52
An orientation reversing path in $M$ would also reverse orientation of the ambient space. – Grumpy Parsnip Feb 13 '12 at 22:53
I'm using that an $n$-dimensional manifold embedded in an $n$-dimensional manifold sits as an open subset, which follows from "invariance of dimension." – Grumpy Parsnip Feb 13 '12 at 22:55
Slightly differently: If there existed an embedding, the image would have to be open by invariance of domain, and open subsets of $\mathbb R^3$ are orientable. – Mariano Suárez-Alvarez Feb 13 '12 at 22:56
There you go :D – Mariano Suárez-Alvarez Feb 13 '12 at 22:58

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