Total linking is intersection with a spanning surface for a link

Choose a basepoint $x_0\in X\setminus F$, which exists on a local side of any interior point of $F$. Represent any class of ${\pi }_1(X,x_0)$ by a continuous based loop, reparametrize it to be constant near the basepoint, and apply relative Whitney approximation followed by relative transversality from [F2]. This gives a smooth based representative transverse to $\mathrm{Int}F$, still constant near $x_0$. Define $I_F$ on that class to be its signed intersection count. If two such loops are homotopic in $X$, reparametrize the homotopy to be constant on endpoint collars. It is then smooth on a neighbourhood of the closed collars, so [F2] smooths it relative to a smaller collar and makes it transverse to $\mathrm{Int}F$. The inverse image of $\mathrm{Int}F$ is a compact oriented one-manifold in the homotopy annulus; it is compact because the annulus image is disjoint from $L=\mathrm{\partial }F$. Orient it using the orientations of the annulus and the normal of $F$. Its oriented boundary signs at the two ends are the intersection signs of the endpoint loops with opposite signs on the two boundary circles. The total signed boundary count is zero, so the endpoint intersection counts agree. Concatenation adds counts, hence $I_F:{\pi }_1(X,x_0)\to \mathbb{Z}$ is a homomorphism.

Near a point of $K_i$, choose oriented local coordinates $(x,y,z)$ in which $K_i$ is the $x$-axis and $F$ is the half-plane $y\ge 0$, $z=0$. The boundary orientation and the stated positive-meridian convention make $(x,y,z)$ ambient-positive. The small positive meridian crosses this half-plane exactly once, at $(0,\rho ,0)$ for some $\rho >0$, in the $+z$ direction. Thus ${\bar{I}}_F([{\mu }_i])=1$ for every $i$.

By [F3], this homomorphism factors through a homomorphism ${\bar{I}}_F:H_1(X;\mathbb{Z})\to \mathbb{Z}$. The signed intersection convention in the Statement agrees with this induced map on the class represented by $J$.

By [F1], the classes $[{\mu }_i]$ form a basis of $H_1(X;\mathbb{Z})$. The homomorphisms ${\bar{I}}_F$ and $\phi$ agree on every basis element by step 1.2 and [F1], so they are equal. Evaluating this equality on $[J]$ gives $J\cdot F=\phi ([J])$.