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Given a closed surface in $\mathbb R^3$, is it necessarily an "embedded surface"? I think it is true, but that is just because I can't think of a closed surface for which we cannot construct a smooth parametrization, though of course that is not a valid argument!

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    No, the Klein bottle has a nice immersion in $\mathbb R^3$ but has no embeddings. Perhaps more to your liking, it is easy to start with an ordinary torus, narrow it partway along and cause that to pass through itself twice, the result resembling a figure 8, but with one narrow tube passing through a fatter tube. http://en.wikipedia.org/wiki/Immersion_%28mathematics%29 and http://en.wikipedia.org/wiki/Wente_torus2012-04-29
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    @WillJagy, thank you!2012-04-30
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    You may look at this link: [Boy's surface](http://en.wikipedia.org/wiki/Boy%27s_surface) is an immersion of $\mathbb{R}P^2$ into $\mathbb{R}^3$ but not embedding. Moreover, any closed non-orientable surface admit no embeddings in $\mathbb{R}^3$.2012-04-30
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    @jerrysciencemath, Thanks!2012-04-30

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