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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!

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  • $\begingroup$ Simple-connectedness is used to establish that $\pi_2(M)\cong H_2(M)$ via Hurewicz. $\endgroup$
    – Neal
    Commented May 17, 2013 at 13:49
  • $\begingroup$ so what? what is the implication of your observation? $\endgroup$
    – fritz
    Commented May 17, 2013 at 15:17

1 Answer 1

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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.

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  • $\begingroup$ wow! I think this is a correct answer! I wonder why hatcher gave no details of this fact. thanks! $\endgroup$
    – fritz
    Commented May 17, 2013 at 21:41
  • $\begingroup$ I think it's correct too. It's also "intuitively obvious", so that may be why Hatcher didn't give details. Also, there may be an easier proof. $\endgroup$ Commented May 18, 2013 at 7:07

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