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It is known that the unknot is the only knot whose complement has fundamental group $\mathbb{Z}$.

Does this fact generalize to links? That is, suppose that $\ell= \ell_1 \cup \dots \cup \ell_n$ is an $n$-component link in $\mathbb{R}^3$ such that $\pi_1(\mathbb{R}^3-\ell)$ is a free group on $n$-generators. Is $\ell$ the unlink on $n$-components?

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A useful result here is

Theorem. (Kneser's conjecture) If $M$ is a compact orientable $3$-manifold with incompressible boundary, and if $\pi_1(M)\approx G_1*G_2$, then there is a connect sum decomposition $M=M_1\mathbin{\#}M_2$ with $\pi_1(M_i)\approx G_i$ for $i=1,2$.

This paper suggests Epstein, "Free products with amalgamation and 3-manifolds", 1961, and Hempel, section 7 for details. (Calegari's notes, Theorem 3.9, has the case of $M$ closed.)

If $L$ is a link with $M=S^3-\nu(L)$ having a compressible boundary, then one of the components is an unknot split from the rest of the link. This is because the loop theorem gives an embedded disk whose boundary is a link component, and the boundary of a regular neighborhood of this disk is a splitting sphere.

Otherwise, if $S^3-\nu(L)$ has incompressible boundary and $\pi_1(S^3-\nu(L))=G_1*G_2$ with each $G_i$ nontrivial, then Kneser's conjecture gives an embedded sphere that splits $L$ into two sublinks.

Recall that $H_1(S^3-L)= \mathbb{Z}^n$, where $n$ is the number of components of $n$. If $\pi_1(S^3-L)$ is a free group of $n$ generators, then this paired with the above implies that $L$ is a completely split link, where each component $L_i$ has $\pi_1(S^3-L_i)=\mathbb{Z}$. The only knot whose knot group is $\mathbb{Z}$ is the unknot, hence each $L_i$ is an unknot.

So, yes, if $\pi_1(S^3-L)$ is a free group on $n$ generators, then $L$ is an $n$-component unlink.

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    $\begingroup$ In fact, the only compact 3-manifolds with nonabelian free fundamental group are the handlebodies; this follows from the sphere theorem. $\endgroup$
    – Hempelicious
    Mar 19, 2020 at 5:06
  • $\begingroup$ @Hempelicious: That claim is false: You need to assume, in addition, that the manifold is irreducible. The complement to an unlink of $>1$ components provides a counter-example to your claim. $\endgroup$
    – Moishe Kohan
    Mar 19, 2020 at 15:31
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    $\begingroup$ @Hempelicious A handlebody is a compact $3$-manifold that, after enough compressions, is a disjoint union of balls. One can show handlebodies are irreducible, but nontrivial connect sums are not irreducible (since they aren't prime). $\endgroup$
    – Kyle Miller
    Mar 23, 2020 at 0:48
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    $\begingroup$ There's also an alternative argument in [Hillman: Alexander Ideals of Links (1981)]: Looking at the abelian subgroup coming from a meridian-longitude pair, this has to be cyclic since by assumption it is a subgroup of a free group. Looking at the image in homology, we see that this is only possible if the longitude is nullhomotopic, and then the loop theorem gives us an embedded disk spanning this link component. So now we can split off the component and continue inductively to recognize the unlink. $\endgroup$
    – ben300694
    Jul 7, 2020 at 10:23
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    $\begingroup$ @user113715 Yes, this is called a peripheral system for the link. (By the way, a link exterior along with a peripheral system is enough to recover the original link. Actually, you just need an exterior along with a meridian curve per boundary component, since gluing thickened disks into the boundary along these curves yields the complement of a disjoint balls in $S^3$, and there is only one way to fill those balls in up to isotopy. If you forget the meridians, then, for example, the Whitehead link cannot be recovered unambiguously.) $\endgroup$
    – Kyle Miller
    Jul 12, 2020 at 20:21

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