# embedding of closed 3-manifolds

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.

-
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. $D^2 X S^1$). – Srikanth Nov 3 '10 at 1:21
A 3-manifold without boundary embedded in $\mathbb{R}^3$ is embedded as an open subset of $\mathbb{R}^3$, hence is not compact. – Jonas Meyer Nov 3 '10 at 1:23

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?