Here exercise 1.2.22 is recalled, but the asker seems to know how to solve it assuming the "geometry is valid". I however, do not know to use the van Kampen theorem in order to find the relations $x_ix_jx^{-1}_i=x_k$ that Hatcher describes. I would like some assistance with this – my geometric intuition is pretty bad.
[Math] Hatcher’s exercise 1.2.22 on the Wirtinger presentation
algebraic-topologyfundamental-groupsgeneral-topologyknot-theory
Related Solutions
This depends on what you mean by being "specifically constructed for this purpose". A nice example (that I'm sure inspired the example given by Hatcher) is the comb space $$X=[0,1]\times\{0\}\cup\{0\}\times [0,1]\cup\bigcup_{n=1}^\infty \left(\left\{\frac{1}{n}\right\}\times [0,1]\right)$$
Any connected subset of the leftmost tooth $\{0\}\times[0,1]$ is a deformation retract of $X$, since $X$ contracts to the point $(0,0)$. But most of the time these won't be strong deformation retracts; in fact the only subset of that tooth that is a strong deformation retract is the point $(0,0)$.
I just consider the case $h=3$, but the argument is completely general.
From the classification of closed surfaces, we know that $N_3$ is the connected sum of three projective spaces, so that $N_3$ be the quotient of an hexagon $P$ by identifying its sides as indicated by the following figure:
Now let $U, V \subset N_3$ be subspaces illustrated by the figure below:
Applying van Kampen theorem, we deduce that $\pi_1(N_3)$ is the quotient of the free product $\pi_1(U) \ast \pi_1(V)$ identifying the images of $\pi_1(U \cap V)$ into $\pi_1(U)$ and $\pi_1(V)$. In fact, $\pi_1(V)$ is clearly trivial so that $\pi_1(N_3)$ turns out to be the quotient of $\pi_1(U)$ by its subgroup corresponding to $\pi_1(U \cap V)$.
Now, there is naturally a deformation retract $\phi$ from $U$ to the quotient of $\partial P$: it is just a graph $\Gamma$, more precisely a bouquet of three circles labelled by $a_1$, $a_2$ and $a_3$. Therefore, $\pi_1(U)$ is just the free group $\langle a_1,a_2,a_3 \mid \ \rangle$.
Then, there is a deformation retract from $U \cap V$ to the circle $\partial V$. Therefore, $\pi_1(U \cap V)$ is infinite cyclic and its image into $\pi_1(U)$ is the homotopy class of the circle $\phi(\partial V)$ in the quotient $\Gamma$ of $\partial P$. Such a loop in $\Gamma$ is labelled by $a_1^2a_2^2a_3^2$.
Finally, we conclude that $$\pi_1(N_3)= \langle a_1,a_2,a_3 \mid a_1^2a_2^2a_3^2 \rangle.$$
Best Answer
Here is a picture taken from Out of Line "Paths and Knot Spaces"
There is more discussion at Topology and Groupoids p,350, and this simple crossing diagram in some sense assumes one is using the fundamental groupoid: insistence on one base points is not natural to the knot situation.
I have demonstrated the crossing relation to children using a copper pentoil and rope, and ended up with this string wrapping on the pentoil:
and asking one of the children to show how the loop comes off the knot!