Omitting the last relation in the Wirtinger presentation of a link group

Question

In my knot theory class homework I encountered the following question:

Prove that for every link, when calculating the Wirtinger presentation of the fundamental group of its complement, you can always omit the last relation in the list.

I tried looking at the last crossing (in a given numbering of the crossings) and find another relation(s) that involves its three generators. It seems - although I can't find a way to prove it - that they always have to appear in other relations too, but I can't seem to find a method to express them. Can anyone help?

Answer 1

This statement is given often without a proof. For example in https://ncatlab.org/nlab/show/knot+group#the_wirtinger_presentation we can read Hint: If doing an example, do not throw away a crossing relation just because it is redundant.. Moreover the same question has been already asked on Math Overflow: https://mathoverflow.net/questions/251292/why-is-any-one-wirtinger-relation-a-consequence-of-the-remainder (and probably more than just one time) but I couldn't find any answers so let's write one :)

Proof from Rolfsen's Knots and links (pages 56-60):

The group $\pi_1(\mathbb R^3 \setminus K)$ is generated by the homotopy classes of the $x_i$ and has presentation $\pi_1(\mathbb R^3 \setminus K) = \langle x_1, \ldots, x_n | r_1 \ldots, r_n \rangle$. Moreover, any one of the $r_i$ may be omitted and the above remains true.

Recall that $K$ lies in the plane $P = \{z = 0\}$ of $\mathbb R^3$, except where it dips down by a distance $\varepsilon$ at each crossing. In order to apply Van Kampen's theorem, we dissect $X := \mathbb R^3 \setminus K$ into $n+2$ pieces $A, B_1, \ldots, B_n$ and $C$. Let $A = \{z \ge - \varepsilon\} \setminus K$. The lower boundary of $A$ is the plane $P' = \{z = -\varepsilon\}$ with $n$ line segments $\beta_1, \ldots, \beta_n$ removed. Let $B_i$ be a solid rectangular box whose top fits on $P'$ and surrounds $\beta_i$. But we remove $\beta_i$ itself from $B_i$, and (in order that $B_i$ contains $*$) adjoin an arc running from the top, straight to $*$, missing $K$. The $B_i$ may be taken to be disjoint from one another. Finally, let $C = $ the closure of everything below $A \cup B_1 \cup \ldots \cup B_n$, plus an arc to $*$.

(...)

To see this, work in $S^3 = \mathbb R^3 \cup \{\infty\}$. Let $A' = A \cup \{\infty\}$ and $C' = B_n \cup C \cup \{\infty\}$. It is clear that $A' \cup B_1 \cup \ldots \cup B_{n-1} \cup C' = S^3 \setminus K$, $\pi_1(A') = \pi(A)$ and adjoining $B_1, \ldots, B_{n-1}$ has the same effect as before. But now we note that $C' \cap (A' \cup B_1 \cup \ldots \cup B_{n-1})$ is simply connected, being a 2-sphere minus an arc, and so is $C'$. Thus we reach the same conclusion without adjoining the relation $r_n$.

There is a picture on page 59 if you have problems imagining the above. In particular, $*$ is a base point above the knot diagram where all loops (elements of $\pi_1$) start and end.