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I am reading the paper,ON ATTACHING 3-HANDLES TO A 1-CONNECTED 4-MANIFOLD by BRUCE TRACE here.He says in this paper that we need only construct a knot $K\subset \partial W^4$ which meets $Σ ^2$ transversely in a single point (i.e., $K$ and $Σ ^2$ are complementary in $\partial W ^4$ ) ,where $W^4$ is 1-connected smooth 4-manifold and $Σ^2$ is 2-sphere in $\partial W^4$.I read this and I can't understand this because I think that $K$ have at least two points in which $K$ intersects $\ Σ^2$. If you can understand why we take such a knot $K$ and find the point of my misunderstanding this ,could you teach me the reason for the existence of $K$ and correct me?

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1 Answer 1

I haven't read the paper, but I assume the sphere $\Sigma^2$ does not bound a $3$-ball, and that the $3$-manifold $\partial W$ is prime: it is not a connected sum of other $3$-manifolds. In this case, the sphere $\Sigma^2$ does not disconnect $\partial W$ into two pieces. So if you take a small transverse arc piercing the $2$-sphere once, the two ends of the arc lie in the same connected component of $\partial W\setminus \Sigma^2$ and can therefore be connected by another arc, which is embedded by transversality.

This happens a lot in $3$-manifolds. The easiest example is $S^1\times S^2$. If $(p,q)\in S^1\times S^2$, then $S^1\times\{q\}$ is a knot intersecting the sphere $\{p\}\times S^2$ exactly once.

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Sorry,I can't understand the definition of prime so,could you give me some examples of prime set? –  Takahiro Oba Jun 30 '12 at 6:46
    
A connected sum of two manifolds is defined by removing a small open ball from each and gluing them together along their spherical boundaries. In particular any $S^{n-1}$ which is embedded in an $n$-manifold and which separates the manifold into two pieces, indicates that the manifold is a nontrivial connected sum of two simpler manifolds. –  Grumpy Parsnip Jun 30 '12 at 7:52
    
OK,I see.Judging from your answer,you say that we can choose such a $K$ only if $\partial W$ is prime,don't you? –  Takahiro Oba Jun 30 '12 at 8:27
    
If you have a nontrivial embedded sphere in a prime manifold, then there is definitely a knot $K$ which hits it one point. In a non-prime manifold, some spheres may have such a $K$ and some may not. –  Grumpy Parsnip Jun 30 '12 at 8:42
    
OK,thank you :) –  Takahiro Oba Jun 30 '12 at 8:50

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