# S^3 (the most basic prime manifold) is prime

So a while ago, I was reading Hatcher’s notes on 3-manifolds. In there, he defines what it means for a manifold to be prime and states, casually, that the 3-sphere $S^3$ is prime. He later says that it follows immediately from Alexander’s Theorem as, and I quote: Every 2-sphere in $S^3$ bounds a 3-ball. And that’s it. Done.

Wait, what?!

Elsewhere, Hatcher expands his above statement: …every 2-sphere in $S^3$ bounds a ball on each side…[and h]ence $S^3$ is prime. Again, though, it isn’t accompanied by anything, and while this is clearly a trivial result, I just couldn’t see it for the longest time…I knew that it followed from a number of things, e.g. the fact that $S^3$ is the identity of the connected sum operation, that $S^3$ is irreducible (and that every irreducible manifold is prime), that one gets the trivial sum $M\# S^3=M$ by splitting along a 2-sphere $S$ in $M^3$ which bounds a 3-ball in $M$, etc. Even so, I didn’t want to leverage some enormous machinery to deduce the smallest of results and what I really wanted was for someone to tell me what I was missing. So I never stopped thinking about this, even after moving forward, until finally – it just clicked!

I figure other people who are as visualization-impaired as I may benefit from seeing this explained in greater depth, so in lieu of typing a blog post containing something new and attention-worthy, I figure I’d share this instead. Details after the break.