A knot complement in $S^3$ has the homology of a circle.Suggest a slogan
Let \(K\subseteq S^3\) be a tame knot, for example a smooth or polygonal one. Then \[ H_i(S^3 \sm K;\ZZ) \cong \begin{cases} \ZZ & i = 0, 1,\\ 0 & \text{otherwise} \end{cases} \] [@Hat02].