TITLE: Constructing 4-manifolds with fundamental group with a given presentation.
QUESTION [13 upvotes]: Reading Princeton Companion I found out that every finitely presented group can be realized as the fundamental group of a 4-manifold. 
When starting to write this answer I found this related MO question. However, my question has to do with one of its answers (which is similar to the hint given in the Princeton Companion). 
The proceedure for constructing this manifold from a given presentation is first to construct a CW-complex with that fundamental group (by wedge sum of circles puting 2-cells to cover the relations) embedding it in $R^5$ and considering the frontier of a tubular neighborhood. 
Two questions come up to me (which are maybe trivial):
1- Why this cannot be done in $R^4$? Or can it be but the result is not the same?
2- Why the resulting manifold has the desired fundamental group?

REPLY [14 votes]: Taking the boundary of tubular neighborhood in codimension two doesn't preserve fundamental group -- the codimension must be at least three.
For a simple example, if you take the boundary of a tubular neighborhood of a single point in $R^2$, you get a circle, which has nontrivial fundamental group.  Similarly, the boundary of a tubular neighborhood of a 2-plane in $R^4$ is the product of a 2-plane with a circle, which again has nontrivial fundamental group.  In general, if you start with a 2-manifold in $R^4$, the boundary of a tubular neighborhood will be a circle bundle over the manifold, and will therefore not have the right fundamental group.
This problem is fixed in $R^5$, because you get a sphere bundle instead of a circle bundle, and the 2-sphere has trivial fundamental group.