I'm completely aware of the triviality of this question, but for some reason, I can't visualize the argument.
In Hatcher's 3-manifold notes, the form of Alexander's theorem stating that Every embedded 2-sphere in $\mathbb{R}^3$ bounds an embedded 3-ball is given. Later, he uses this fact to conclude that $S^3$ is prime (recall that a 3-manifold $M$ is prime if, whenever $M=P\#Q$, either $P=S^3$ or $Q=S^3$ where here, $P\#Q$ denotes the connected sum of $P$ and $Q$), stating merely that every 2-sphere in S^3 bounds a 3-ball. I was hoping to visualize why this is true, and so far, I'm not having any luck; I'm hoping someone can help.
Things I've read:
- From Hatcher's notes, it's mentioned that the trivial decomposition $M=M\# S^3$ is obtained by choosing the sphere $S$ (in the connected sum decomposition process) which bounds a ball in $M$. I realize that knowing this implies my result immediately, but it doesn't help me see the result.
- Elsewhere, Hatcher states that the result follows from the fact that every 2-sphere in $S^3$ bounds a ball on each side. I've seen this justification elsewhere as well, but I can't visualize this one any easier.
I guess what I'm looking for is a direct proof of some kind. I tried to construct one as follows, but I'm stuck almost immediately after doing all the obvious things. Note that for a sphere $S$ in $M$, $M|S$ is Hatcher's notation for the manifold $M$ obtained by splitting along $S$, i.e. by removing an open tubular neighborhood $N(S)$ of $S$ from $M$.
Suppose that $S^3=P\#Q$. By definition of connected sum, there exists a sphere $S$ in $S^3$ such that $S^3|S$ has two components $P'$ and $Q'$ where $P$, respectively $Q$, is obtained from $P'$, respectively $Q'$, by filling in the boundary sphere corresponding to $S$ with a ball. By Alexander's theorem, $S$ bounds a 3-ball $B$ in $S^3$, so....
And that's it. I really have no idea how to proceed, and despite this being truly one of the most trivial things imaginable, I'm at a loss. Knowing why this is true for $S^3$ would be great, but knowing why the fact about general $M^3$ being trivially decomposed when $S$ bounds a ball would probably be much more helpful from a big picture perspective.
Imagine an open ball $B^2$ in $\mathbb R^2$. If you consider the one point compactification of the complement, you get a closed ball. You could also consider an open ball in the one point compactification of $R^2$ which is $S^2$. Take the ball out and you'll see another ball.
The reason for this is that the boundary of the embedded ball, namely $S^1 = \partial B^2$, closes up at the other end, namely $\infty$.
This is how you can image the higher-dimensional analogue. The boundary of the ball, is also the boundary of another ball.
You could also imagine a bicollar of a codimension 1 sphere - it will close up at both ends.
Hope it helps your intuition.