A knot exterior has first homology generated by its meridian

The solid torus $N(K)=S^1\times D^2$ deformation retracts to its core circle, so by [F4] it has $H_1(N(K);\mathbb{Z})\cong \mathbb{Z}$ and $H_2(N(K);\mathbb{Z})=0$. The product torus has $H_1\cong {\mathbb{Z}}^2$ by [F2]; inclusion sends the product section to the core generator and the meridian to zero because the meridian bounds a fiber disk. Hence $H_1(\mathrm{\partial }N(K))\to H_1(N(K))$ is surjective with kernel exactly $\mathbb{Z}[{\mu }_K]$. Exactness in [F3] identifies $H_2(N(K),\mathrm{\partial }N(K);\mathbb{Z})$ with $\mathbb{Z}$, and its boundary map sends a relative normal disk generator to $[{\mu }_K]$.

Enlarge $N(K)$ by a product collar and apply excision [F3] to the pair $(S^3,E_K)$. It identifies $H_2(S^3,E_K;\mathbb{Z})$ with $H_2(N(K),\mathrm{\partial }N(K);\mathbb{Z})$, with the normal disk generator preserved. By the long exact sequence of $(S^3,E_K)$ and [F4], the connecting map $H_2(S^3,E_K)\to H_1(E_K)$ is an isomorphism, because the adjacent groups $H_2(S^3)$ and $H_1(S^3)$ vanish. Naturality sends the relative normal disk to its boundary positive meridian. Thus $H_1(E_K;\mathbb{Z})$ is infinite cyclic on $[{\mu }_K]$.