# Is a compact, simply-connected 3-manifold necessarily $S^3$ with $B^3$'s removed?

Let $M$ be a compact, simply-connected 3-manifold (which is also smooth and connected). Is $M$ diffeomorphic to $S^3$ with a finite number of $B^3$'s removed?

This seems like a handy fact, but I haven't ever come across it.

My reasoning: If $\partial M = \emptyset$, then $M$ is $S^3$ by Perelman's theorem (Poincare Conjecture). Otherwise, $\partial M$ is a finite union of closed surfaces. Simple-connectedness implies $M$ (and thus $\partial M$) is orientable. Simple-connectedness also implies that the map on $H_1$ induced by the inclusion $i:\partial M \hookrightarrow M$ is the zero map, so "half lives, half dies" gives $$\operatorname{rank}(H_1(\partial M))=2\operatorname{rank}(\ker i_*)=2 \operatorname{rank}(H_1(\partial M)).$$ Therefore $H_1(\partial M)=0$, implying that $\partial M$ is a union of 2-spheres. Capping off these boundary components with 3-balls (which preserves the fundamental group), we obtain a compact, simply-connected 3-manifold $M'$ which is closed, hence diffeomorphic to $S^3$. Thus $M$ is diffeomorphic to $S^3$ with a finite number of 3-balls removed.

• In general dimensions, gluing balls into spheres doesn't give you diffeomorphisms, only homeomorphisms (Alexander trick). But in dim 3, homeo and diffeo classifications are the same. What is this "half lives, half dies" thing? (I am not a 3-manifold topologist) Commented Apr 8, 2015 at 15:32
• @JasonDeVito: If $M$ is any compact, orientable 3-manifold, then $\partial M$ is a closed, orientable surface, so $H_1(\partial M)$ has even rank. The map induced by the inclusion $\partial M \hookrightarrow M$ "kills" half of $H_1(\partial M)$, while the other half "lives". In a solid torus, for example, the longitude in the boundary is also homologically nontrivial in the solid torus, but the meridian bounds a disk and "dies".
– Kyle
Commented Apr 8, 2015 at 15:42
• @JasonDeVito: And gluing in $B^3$'s is the part I'm most concerned about anyway, so the equivalence of homeo/diffeo classifications helps with the sanity check.
– Kyle
Commented Apr 8, 2015 at 15:43
• The "half-lives, half-dies" thing is a consequence of the long exact sequence of homology rel boundary coupled with Poincare duality rel boundary. Commented Apr 8, 2015 at 17:24
• easy fact: by exactness $0\cong H^1(M) \cong H_2(M,\partial M) \to H_1(\partial M) \to H_1(M) \cong 0$, the boundary surface has positive Euler characteristic. Commented Apr 8, 2015 at 18:35

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 $\partial M=S^2 \sqcup \cdots \sqcup S^2$:
1. "Half lives, half dies" shows $H_1(\partial M)=0$, hence $\partial M$ consists of copies of $S^2$.
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.