# Does the abelianization of the fundamental group of this knot-complement really $\mathbb{Z}$?

I have a doubt about some edge cases in problem $$22.(b)$$ in Hatcher's "Algebraic Topology", section $$1.2$$.

The writer presents a knot:

To begin, we position the knot to lie almost flat on a table, so that $$K$$ consists of finitely many disjoint arcs $$\alpha_i$$ where it intersects the table top together with finitely many disjoint arcs $$\beta_\ell$$ where $$K$$ crosses over itself.

The configuration at such a crossing is shown in the first figure below. The writer than proceeds to describe a cell-complex that $$\mathbb{R}^3 -K$$ deformation-retracts onto, and that will help us calculate $$\pi_{1}(\mathbb{R}^3 -K)$$. See the essentials bellow:

Start with the rectangle $$T$$ formed by the table top. Above each arc $$\alpha_i$$ place a rectangular strip $$R_i$$, curved to run parallel to $$\alpha_i$$ and arched so that the two long edges of $$R_i$$ are identified with points of $$T$$ , as in the second figure.

Any arcs $$\beta_\ell$$ that cross over $$\alpha_i$$ are positioned to lie in $$R_i$$.

Finally, over each arc $$\beta_\ell$$ put a square $$S_\ell$$ , bent downward along its four edges so that these edges are identified with points of three strips $$R_i, R_j$$ and $$R_k$$ as in the third figure; namely, two opposite edges of $$S_\ell$$ are identified with short edges of $$R_j$$ and $$R_k$$ and the other two opposite edges of Sℓ are identified with two arcs crossing the interior of $$R_i$$ .

The knot K is now a subspace of $$X$$ , but after we lift $$K$$ up slightly into the complement of $$X$$ , it becomes evident that $$X$$ is a deformation retract of $$\mathbb{R}^3 -K$$.

I've managed to calculate the fundamental group, which is described bellow for completeness:

$$\pi_{1}(\mathbb{R}^3 -K)$$ has a presentation with one generator $$x_i$$ for each strip $$R_i$$ and one relation of the form $$x_i x_j x^{−1}_i = x_k$$ for each square $$S_\ell$$ , where the indices are as in the figures above.

Than I run into some problems - We're asked to:

Use this presentation to show that the abelianization of $$\pi_{1}(\mathbb{R}^3 -K)$$ is $$\mathbb{Z}$$.

I don't think this is correct. It is right for the exact configuration sketched in the figure, but it won't work if the set of arcs labeled as $$\{\beta_l\}$$ is empty, and there are at least 2 different $$\alpha$$-type arcs. Such a configuration deformation retracts to $$S^1 \vee S^1$$, and the abelianization in that case is $$\mathbb{Z} \times \mathbb{Z}$$

If I'm not mistaken, this won't work even with the assumption that $$\{\beta_l\}\neq \emptyset$$, which can be seen when taking two parallel copies of the intersection described: This violate the desired corollary, since if two loops are homotopic in $$\mathbb{R}^3 -K$$, they are obviously homotopic in $$\mathbb{R}^3 - (\alpha_i \cup \{red\ arc\ \})$$ and this, again, deformation retracts into $$S^1 \vee S^1$$, which leads into a contradiction.

What did I miss?

• It is a general theorem which has many different proofs that the 1st homology of a knot complement in $R^3$ is ${\mathbb Z}$. Apr 23 at 14:21
• I know very little about knots - Just read wikipedia's definition; Wiki says that a knot is an embedding of $S^1 \hookrightarrow \mathbb{R}^3$. This thing above doesn't seem homeomorphic to the sphere.
– NG_
Apr 23 at 14:36

If you accept that $$\pi_{1}(\mathbb{R}^3 -K)$$ has a presentation with one generator $$x_i$$ for each strip $$R_i$$ and one relation of the form $$x_i x_j x^{−1}_i = x_k$$ for each square $$S_\ell$$, then you see that the abelianization has generators $$x_i$$ and relations $$x_i x_j x^{−1}_i = x_k$$ plus $$x_i x_j = x_j x_i$$. Thus $$x_j = x_jx_ix_i^{-1} = x_i x_j x^{−1}_i = x_k$$ which shows that all generators $$x_i$$ are identified to a single generator (which gives an infinite cyclic group).
He splits $$K$$ into disjoint arcs $$\alpha_i$$ and $$\beta_l$$. This splitting is determined by the set $$C$$ of crossing points of $$K$$. What happens if there are no crossing points? Formally the approach breaks down, but we can rescue it. We can add finitely many points of $$K$$ to $$C$$ and obtain a set $$C' \supset C$$. This gives again a splitting into disjoint arcs $$\alpha_i$$ and $$\beta_l$$. At each crossing point we get relations of the form $$x_i x_j x^{−1}_i = x_k$$. At the non-crossing points two segments $$\alpha_j$$ and $$\alpha_k$$ meet, but there is no $$\beta_i$$ producing a relation. Formally we get a relation $$ex_je^{-1} = x_k$$ (i.e. $$x_j = x_k$$), where $$e$$ is neutral element of the fundamental group.
Doing this for a knot $$K$$ without crossing point, we can split $$K$$ into disjoint arcs $$\alpha_i$$ and get the relations $$x_j = x_k$$ at all points of the set $$C'$$. Thus we end up with one generator which gives an infinite cyclic group (before abelianization).
• I think my real issue is the summary for the first paragraph: "...which shows that all generators are identified ...". I think I can prove it under the assumption that $K$ is an embedding for $S^1$ - which I did not assume at first. After reading a bit more I realize that this is a necessity BTW, I think this assumption also eliminates the need for adding more points to C, since the only case with no crossing will have just one rectangle - This provides us with the desired fundamental group anyway.