Knot complements in $S^3$ are aspherical, and deleting a nullhomologous knot contributes a meridional $\ZZ$ to first homology.Suggest a slogan
Let \(K\subseteq S^3\) be a knot. Then \(S^3\sm K\) is a \(K(\pi, 1)\) [@Hat02], and \(\RR^3 \sm K \homotopic S^2 \vee \qty{S^3 \sm K}\). If \(X\) is an orientable \(3\)-manifold and \(K\subseteq X\) is a nullhomologous tame knot with tubular neighborhood \(\nu(K)\), then \[ H_1\qty{X\sm \nu(K)} \cong H_1(X) \oplus \ZZ ,\] the \(\ZZ\) summand generated by a meridian of \(K\).