# Showing that gluing two knot exteriors together contains subgroups isomorphic with the knot groups.

I'm working through Rolfsen's "Knots and Links" and section 9D exercise 10 has me stumped:

Let $K_1$ and $K_2$ be knots in two separate copies of $S^3$ with respective meridians $m_1$ and $m_2$ and preferred longitudes $l_1$ and $l_2$. Show that if you take the exteriors of (tubular neighborhoods of the) two knots (in $S^3$) and attach them by a homeomorphism along their boundaries such that $h(m_1) = l_2$ and $h(l_1) = m_2$. Show that the fundamental group of the resulting space contains subgroups isomorphic to the knot groups $\pi_1(S^3-K_1)$ and $\pi_1(S^3-K_2)$.

It seems like Van Kampen's theorem is the obvious route, but I can't get a handle on what this homeomorphism is supposed to do to my generators of the knot groups.

EDIT: This problem also requires that $K_1$ and $K_2$ be non-trivial.

• The result is false for trivial knots. If $K_1$ and $K_2$ are trivial knots (so that $\pi_1(S^3 \setminus K_i) \simeq \mathbb Z$) with their usual preferred longitude (given by the Seifert disc, for instance), their exteriors are solid tori and the gluing you describe gives the classical genus $1$ Heegaard splitting $3$-sphere, so that $\pi_1(\text{gluing}) = 0$. – PseudoNeo May 7 '15 at 6:09
• The problem in Rolfsen demands that both of these knots be nontrivial to start with. – user98602 May 7 '15 at 7:04
• Yes, sorry. I was looking at this with some people, and most of our discussion was about the requirement that the knots be non-trivial. And yet, I forgot to put it in this post. I'll edit it to reflect that requirement. – Minirogue May 7 '15 at 23:28

Suppose your two knot exteriors are $M_1$ and $M_2$. van Kampen's theorem says that $\pi_1(M_1 \cup_h M_2)$ is the amalgamated free product of $\pi_1(M_1)$ and $\pi_1(M_2)$, amalgamated over $\pi_1(T^2)$, the fundamental group of their intersection. There is a canonical map $\pi_1(M_1) \to \pi_1(M_1 \cup_h M_2)$ given by inclusion $M_1 \hookrightarrow M_1 \cup_h M_2$; the map this induces on fundamental groups is precisely the canonical map $\pi_1(M_1) \to \pi_1(M_1) *_{\pi_1(T^2)} \pi_1(M_2)$. Let $i_1, i_2$ be the inclusions of $M_1 \cap M_2$ into $M_1$ and $M_2$ respectively. This map is going to have kernel iff some element of $\pi_1(M_1)$ is in the normal closure of $\{i_1^*(a)i_2^*(a)^{-1} : a \in \pi_1(T^2)\}$.
The normal closure is just given by adding in all the conjugates of elements of the subgroup. So suppose $s^{-1}i_1^*(a)i_2^*(a)^{-1}s$ was a word all of whose terms (in the reduced word) are in $\pi_1(M_1)$. If $i_2^*(a)$ is nontrivial, $s$ has to be of the form $i_2^*(a)w$, where $w \in \pi_1(M_1)$. This gives $w^{-1}i_2^*(a)^{-1} i_1^*(a)$, which is no better than how we started; so conjugation can't get rid of the $i_2^*(a)$ if it's nonzero, and no nonzero words of $\pi_1(M)$ are in the kernel if $i_2^*(a)$ is nonzero for every $a$.
It suffices to note, now, that $i_2^*(a)$ is nonzero for every nonzero $a$. But by assumption the knots we're doing this from are knotted, and being knotted is equivalent to the statement that the inclusion of the boundary into the knot exterior induces an injection on fundamental groups. So the canonical map $\pi_1(M_1) \to \pi_1(M_1 \cup_h M_2)$ is an injection. Because of the symmetry of the situation, the same argument works for $\pi_1(M_2)$.