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. enter image description here

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: enter image description here

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?

  • $\begingroup$ It is a general theorem which has many different proofs that the 1st homology of a knot complement in $R^3$ is ${\mathbb Z}$. $\endgroup$ Apr 23 at 14:21
  • $\begingroup$ 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. $\endgroup$
    – NG_
    Apr 23 at 14:36

1 Answer 1


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).

The problem addressed in your question is the transition from the geometric description of a knot to the algebraic representation of the fundamental group of its complement. Hatcher is a bit sloppy here.

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).

  • $\begingroup$ 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. $\endgroup$
    – NG_
    Apr 24 at 6:01

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