# On compact, orientable 3-manifolds with non-empty boundary

I recall my Professor having stated something along the lines of the following, but I am not quite certain about the precise statement she gave:

Let $M$ be a compact, orientable 3 manifold with non-empty boundary. Then $M$ can be embedded in $\mathbb S^3$. More precisely, $M$ is diffeomorphic to $\mathbb S^3 \setminus N$, where $N$ is a finite collection of embedded open handlebodies.

Is this statement true? If not, what would be an easy counterexample, and does there still exists a weaker form of the statement that is actually true ? Any help is appreciated.

• As stated this cannot be correct. If the boundary is a $2$-sphere, then this $2$-sphere separates $S^3$ into two $3$-disks (assuming the embeeding is smooth), so your manifold must be a disk. – Igor Belegradek Jan 2 '16 at 18:35
• Maybe adding the assumption that the boundary surfaces have genus at least $1$ can fix this... – H1ghfiv3 Jan 2 '16 at 18:38
• Possibly you misheard a statement of the Fox re-embedding theorem, which states that if M is a submanifold of S^3 to begin with, you can re-embed it in S^3 with complement being handlebodies. – Chan-Ho Suh Jan 2 '16 at 20:05
• Immersion rather than embedding? – Włodzimierz Holsztyński Jan 3 '16 at 8:44
• I believe it must have been Fox re-embedding theorem. – H1ghfiv3 Jan 3 '16 at 10:03

You can't embed $\mathbb{RP}^2$ in $S^3$ but you can embed it into $\mathbb{RP}^3$ which is orientable. (Take the quotient of $S^3$ by the antipodal map. The equatorial $S^2$ becomes a one-sided projective plane.) Drilling out a torus in a ball from $\mathbb{RP}^3$ doesn't make it embeddable in $S^3$ because any embedding would give an embedding of a disjoint $\mathbb{RP}^2$.
More generally, consider drilling out a handlebody within a ball within some compact manifold that is not $S^3$. If you could embed this, then you could embed the smaller manifold where you remove the whole ball, which Igor Belegradek's comment shows is not possible.
• And what if we additionally assume $M$ to be irreducible ? – H1ghfiv3 Jan 2 '16 at 19:33
• Still false. If you can embed $M$ in $S^3$ then Alexander duality greatly restricts the homology of $M$, and being irreducible doesn't restrict the homology. For example, the product of a punctured surface with $S^1$ typically doesn't embed in $S^3$ and has a ball as a universal cover so it is irreducible. – Douglas Zare Jan 2 '16 at 20:11