Is a compact, simply-connected 3-manifold necessarily $S^3$ with $B^3$'s removed?
Solution 1:
The answer is yes.
Approach 1: As in the original argument, it suffices to show that all boundary components are homeomorphic to $S^2$, since we can then uniquely glue in 3-balls to produce a closed 3-manifold that is still simply-connected, hence diffeomorphic to $S^3$. Here are two ways to show that $\partial M=S^2 \sqcup \cdots \sqcup S^2$:
"Half lives, half dies" shows $H_1(\partial M)=0$, hence $\partial M$ consists of copies of $S^2$.
From Dan: Poincaré-Lefschetz duality and simple-connectedness give $H_2(M,\partial M)\cong H^1(M) \cong 0$, so part of the LES for $(M,\partial M)$ reads $0 \cong H_2(M,\partial M)\to H_1(\partial M)\to H_1(M)\cong 0$. This implies $H_1(\partial M)=0$.
Approach 2: The boundary components of $M$ are closed orientable surfaces, so they can be filled in with 3-dimensional handlebodies. Observe that gluing in handlebodies preserves simple-connectedness regardless of the gluing maps, since any loop in a handlebody $Y$ has a representative in $\partial Y$ that must be nullhomotopic when viewed in $M$. Filling in all boundary components produces a closed simply-connected 3-manifold, i.e. $S^3$. Now we work in reverse: The complement of a handlebody in $S^3$ is simply-connected only if the handlebody has genus zero, i.e. is $B^3$. It follows that $M$ is $S^3$ with $B^3$'s removed.