Infinity cyclic cover for knot complement

Question

I was confronted with this phrase:

Let $L$ be a link in $\mathbb{S}^3$ with complement $X$ and infinite cyclic cover $p:X_{\infty}\rightarrow X.$ Recall that $p_\ast :\pi_1(X_\infty)\to\pi_1(X)$ is an isomorphism onto the commutator subgroup $[\pi_1(X),\pi_1(X)]$.

Now two question arised.

Question 1. Why does a knot complement has an infinity cyclic cover? (An infinity cyclic cover is nothing but a cover with fiber $\mathbb{Z}$, isn't it?)

Question 2. Why is $p_\ast (\pi_1(X_\infty))=[\pi_1(X),\pi_1(X)]$?

EDIT: A construction for $X_\infty$ can be found [here].

Answer 1

You need to use the fact that the abelianization of the knot group is $\mathbb Z$, so that the cover corresponding to the commutator subgroup is actually infinite cyclic.

Edit: Recall that the classification of covering spaces gives a bijection between subgroups of the fundamental group and covering spaces. So there is a cover corresponding to the commutator subgroup. This is a normal subgroup and so the group of deck transformations is $\pi_1(X)/[\pi_1(X),\pi_1(X)]$. It is known that for knots this quotient (called the abelianization) is infinite cyclic. Thus you get a cyclic cover.