Can someone explain for me following sentence "Mike Freedman has proven all homology 3-spheres admit tame topological embeddings into $S^4$.", which I found on "open problem garden" http://www.openproblemgarden.org/op/which_compact_boundaryless_3_manifolds_embed_smoothly_in_the_4_sphere ?
On wikipedia I read that "Surgery on a knot in the 3-sphere with framing +1 or −1 gives homology sphere". From the topology books I read long time ago I remember that $H_1(X)$ is abelianization of $\pi_1(X)$.
My question has two parts 1) understand which 3-manifolds are "homology spheres" 2) how the embedding in $S^4$ for "homology sphere" is constructed ?
I have also asked question on mathoverflow whether exists method to determine which 3-manifold embeds in $R^4$ or $S^4$. I understand that on mathoverflow I should ask advanced questions while on stackexchange I should aks the basic ones $:)$
I add this clarification, because someone asked what is my question here. My original interest was which closed 3-manifolds embeds in $R^4$. I found out on mathoverflow from Ryan Budney that "Poincare homology sphere has tame topological embedding in $S^4$ but it does not have smooth embedding.". This another mistery for me. My intuition is that 3-manifolds can be divided to two families: first one can be embedded to $R^4$. Second one can only be embedded in $R^5$. Next we could make order in these families by some invariants like fundamental group.
To gain some intuition I would like to see some examples of 3-manifolds which can be embedded in $R^4$. One of the examples are "homology spheres". They have trivial homology groups, but they can have finite or infinite fundamental group. Homology spheres embed in $S^4$ with tame topological embedding. According to Ryan Budney explanations: "This is a result of Mike Freedman's. The embedding construction involves a type of infinite handle adjunctions. So the fact that you have a tubular neighbourhood is not obvious at all".
I would like to see also examples of 3-manifolds which cannot be embedded in $R^4$.