question from hatcher basic 3 manifolds The question is: why should a homologically trivial embedded sphere in a simply connected  (not necessarily compact) 3 manifold M bound a compact 3 manifold embedded in M?
I had this problem reading the first part of the proof of proposition 3.10 in "hatcher on 3 manifolds topology", available here:
http://www.math.cornell.edu/~hatcher/3M/3M.pdf page 52

Warning! I don't know if the simple connectedness is necessary. Better is read directly from the link!
 A: Here's an argument that I can see. There may be an easier method. There is an intersection pairing on compact oriented manifolds:
$$H_p(M)\otimes H_{n-p}(M,\partial M)\to \mathbb Z$$
and a similar pairing for unoriented manifolds:
$$H_p(M;\mathbb Z_2)\otimes H_{n-p}(M,\partial M;\mathbb Z_2)\to \mathbb Z_2.$$
These can be described by looking at the algebraic intersection number of representatives of the homology classes. In particular, if you can ever find two manifolds that intersect transversely in a single point, then both represent nontrivial homology classes. Therefore, assuming $M$ is a compact manifold for the time being, if there is a homologically trivial embedded sphere, it must separate $M$ into two pieces (else it meets a circle in a point) and it cannot separate boundary components (else it meets an arc from the boundary to itself in one point.) So it is the boundary of the part of $M$ that it separates off with no boundary components. Okay, so this proves the case for $M$ compact with boundary. Suppose now that $M$ is not compact. Then the homologically zero sphere bounds some singular chain in $M$, which is compact. Now take a tame neighborhood of this singular chain. (That is, take a small neighborhood that has a well-defined 2-dimensional boundary.) Now apply the previous argument to this new manifold.
