Homology and homotopy type for knot complements

2012-06-30T07:00:12Z

I'm trying to understand a paper by Kawauchi and Matumoto, 'An estimate of infinite cyclic coverings and knot theory.' In one of their proofs they have a manifold $E$ which is the complement of a codimension-2 $n$-knot, with its infinite cyclic cover $\widetilde{E}$. They show that $\widetilde{H}_{*}(\widetilde{E};\mathbb{Z})=0$. Then they claim that because of this we must have that $E$ is homotopy equivalent to $S^{1}$.

It is clear that this shows $E$ is homologically equivalent to $S^1$, but why do we also get the result that it is homotopy equivalent?

Answer by Aru Ray for Homology and homotopy type for knot complements

2012-06-30T14:21:12Z

The result being proved is:

Let $n ≥ 3$. A ribbon $n$-knot $K$ is unknotted if $\pi_1(\mathbb{S}^{n+2}−K)\cong \mathbb{Z}$

Let $E$ denote $\mathbb{S}^{n+2}−K$, then $\pi_1(E) \cong \mathbb{Z}$ by hypothesis. Let $\tilde{E}$ denote the infinite cyclic cover of $E$. $\pi_2(E)\cong \pi_2(\tilde{E})\cong H_2(\tilde{E})=0$, since $\tilde{E}$ covers $E$ and by the Hurewicz theorem, since (it is proved in the paper as mentioned by the OP) $\tilde{H}(\tilde{E};\mathbb{Z})\cong 0$ and $\pi_1(\tilde{E})\cong 0$ (as the infinite cyclic cover of a space with $\pi_1 = \mathbb{Z}$). Keep doing this to get that $\pi_1(E)\cong \mathbb{Z}$ and $\pi_i(E)\cong 0$, for all $i>1$. This means that $E$ is a $K(\mathbb{Z},1)$. All of those are homotopy equivalent to each other, so $E$ is homotopy equivalent to $\mathbb{S}^1$

Homology and homotopy type for knot complements - MathOverflow

Homology and homotopy type for knot complements

Answer by Aru Ray for Homology and homotopy type for knot complements