Is the surface bounding the complement of an unlink in 3-space incompressible?

Question

Let $L$ be an unlink of $n>1$ components in $\mathbb{S}^3$. Let $N=\mathbb{S}^3-L$. Is the boundary of $N$ incompressible?

Answer 1

Strictly, the question is malformed. A compressible surface must be compact and properly embedded in the ambient space. If we take $N$ to be the ambient space, its boundary is not a properly embedded surface so is neither compressible nor incompressible. To overcome this objection, we take the compact properly embedded surface, $S$, to be a collar of $\partial N$ -- a boundary parallel copy of the boundary, displaced into $N$ by an amount small enough that $S$ is embedded (not self-intersecting).

Note that when you delete an unknot from $S^3$, the resulting boundary component is a torus, $T^2 \simeq S^1 \times S^1$. So now we ask if $\partial N$, a union of disjoint tori, is compressible.

Pick a torus boundary component, $t$, and pick preferred longitude, $\ell$, of $t$. (The preferred longitude does not link the "missing" solid torus, so is nullhomotopic in $N$. The homotopy exhibiting this traces out a disk in $N$.) This $\ell$ bounds a properly embedded disk in $N$, but does not bound a disk in $t$ (or $S$) so is a nontrivial compressing disk. Since $S$ has a nontrivial compressing disk, it is compressible

An embedding of the properly embedded disk is in red with $t$ in orange. (The grid lines on the torus are families of meridians and preferred longitudes.)

The red disk bounds an annulus, not a disk, in $S$.