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Prove or disprove that closed 3-manifolds which are not simply connected cannot be embedded in three-dimensional Euclidean spaces. I am not a mathematics major and I am taking introductory topology this semester. But I need to apply this result for my research. Any help is much appreciated.

Thanks, Srikanth.

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    I am not sure how to prove this, but I did try to come up with counter-examples. All the three manifolds that are not simply connected that I can embed in $\mathbb{R}^3$ are not closed (they have a boundary e.g. $D^2 X S^1$).2010-11-03
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    A 3-manifold without boundary embedded in $\mathbb{R}^3$ is embedded as an open subset of $\mathbb{R}^3$, hence is not compact.2010-11-03

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If your manifold is closed you must be using some definition that ensures it's compact? So if $f : M \to \mathbb R^3$ is an embedding, isn't the image simultanously compact and open?

edit in response to your comment: if $f : M \to \mathbb R^3$ is an embedding, let $B \subset M$ be an open subset of $M$ which is homeomorphic to an open ball in $\mathbb R^3$. You need to argue that $f(B)$ is open in $\mathbb R^3$. If your embedding is smooth there's a big theorem from calculus that gives you the result. If you're talking about topological embeddings you're going to need a tool. Have you studied the "invariance of dimension" theorem?

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    I am sorry if my question seems a little vague. I am still learning topology but for now my knowledge mostly comes from Wikipedia. The 3-manifolds that I am looking at are all spherical 3-manifolds (http://en.wikipedia.org/wiki/Spherical_3-manifold) and it is mentioned that they are closed. I do agree that they are compact (being closed and bounded) and I understand that the embedding is compact as well (since compactness is a topological property). I don't understand why the embedding is open. Can you please give me some idea. I will try to read more and understand this. Thank you very much.2010-11-03
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    Not yet. But I will read up on it. Thanks a lot. This is really helpful.2010-11-03