The exterior of a slice disk has first homology generated by a meridian

The portion of $\mathrm{\partial }X$ added by deleting $\nu (D)$ is the normal-circle bundle $D^2\times S^1$, and it meets $E_K$ in $S^1\times S^1$. By excision [F2], the relative pair $(B^4,X)$ has the homology of $(D^2\times D^2,D^2\times S^1)$. The latter solid-torus subspace has $H_1\cong \mathbb{Z}$ generated by a normal circle and $H_2=0$, while $D^2\times D^2$ is contractible by [F3]. Its pair sequence [F2] therefore identifies $H_2(B^4,X;\mathbb{Z})$ with $\mathbb{Z}$, generated by a normal disk whose boundary is that normal circle.

In the pair sequence of $(B^4,X)$, the segment $H_2(B^4)⟶H_2(B^4,X)\stackrel{\mathrm{\partial }}{\to }{H}_1(X)⟶H_1(B^4)$ has zero end groups by [F3]. Hence $\mathrm{\partial }$ is an isomorphism. The generator from step 1.1 maps to the normal circle on $D^2\times S^1$. A path in the base disk $D^2$ from an interior point to a boundary point gives a homotopy of these normal circles to the boundary meridian of $K$, with orientation fixed by [F4]. Thus $H_1(X;\mathbb{Z})=\mathbb{Z}[{\mu }_X]$.

The inclusion $E_K\hookrightarrow X$ carries the boundary meridian ${\mu }_K$ to the same normal circle ${\mu }_X$. It is therefore an isomorphism between the infinite cyclic groups in [F1] and step 2.1.