Poincaré dual of the Alexander dual of the fundamental class of a knot is given by a Seifert surface Let $K\subset S^3$ be an oriented knot and let $F:\overline{B^2}\times K\rightarrow S^3$ be a thickening with self linking number $0$. I will denote $F(B^2\times K)$ by $(B^2\times K)$ for simplicity. Write $X_K:=S^3\setminus(B^2\times K)$ for the knot complement and denote by $i:\partial X_K\rightarrow X$ the canonical inclusion. Denote by $[\mu_K^\star]\in H^1(S^3\setminus(B^2\times K)$ the unique generator with the property that $\langle[\mu_K^\star],\mu_K\rangle =1$ for $\mu_K$ a meridian of $K$. In other words $[\mu_K^\star]$ is the Alexander dual of the fundamental class of $K$. Further let $F_K$ be a Seifert surface for $K$.
I want to show that
$$\left[ \mu_K^\star\right] \frown \left[ S^3\setminus(B^2\times K)\right] = i^J_\star(\left[ F_K'\right]),$$
where $F^\prime_K:=X_K\cap F_K$. To do so I already have proven that the boundary map
$$\partial_2:H_2(S^3\setminus(B^2\times K),S^1\times J;\mathbb{Z})\rightarrow H_1(S^1\times K;\mathbb{Z})$$
is a monomorphism thus it suffices to show that
$$\partial_2 (\left[ \mu_K^\star\right] \frown \left[ S^3\setminus(B^2\times K)\right] )=\partial_2 (\left[ F_K'\right] ).$$
In particular it suffices to show that
$$\partial([\mu_K^\star]\frown[S^3\setminus (B^2\times K)]) = [\lambda_K],$$
where $\lambda_K$ denotes a longitude for $K$.
I started with using the almost naturality of the Poincaré Duality with respect to the long exact sequence of a pair, this is that the square
$\require{AMScd}$
\begin{CD}
H^k(S^3\setminus(B^2\times K);\mathbb{Z}) @>i^\star>> H^{k}(S^1\times K;\mathbb{Z})\\
@VV\frown[S^3\setminus(B^2\times K]V @V \frown[S^1\times K]VV\\
H_{n-k}(S^3\setminus(B^2\times K),S^1\times K;\mathbb{Z} @>\partial_{n-k}>> H_{n-k-1}(S^1\times K;\mathbb{Z})\\
\end{CD}
commutes up to $(-1)^k$.
Thus one could start calculating
$$\partial_2(\left[ \mu_K^\star\right] \frown\left[ S^3\setminus F(B^2\times K)\right] )        \overset{\text{above}}{=}       -i^\star(\left[ \mu_K^\star\right] )\frown\left[ \partial(S^3\setminus F(B^2\times K))\right]=\color{red}{???}=[\lambda_K]$$
It smells to me like we need to use the
property
$$\varphi\frown f_\star(\sigma)=f_\star(f^\star(\varphi)\frown\sigma)$$ with respect to our thickening. I.p. as Kyle Miller pointed out to me in another question we can rewrite the above to
$$\partial_2(\left[ \mu_K^\star\right] \frown\left[ S^3\setminus F(B^2\times K)\right] )        \overset{\text{above}}{=}       -i^\star(\left[ \mu_K^\star\right] )\frown\left[ \partial(S^3\setminus F(B^2\times K))\right]=$$
$$=-i^\star(\left[ \mu_K^\star\right] )\frown(-F_\star[S^1\times K])=\color{red}{???}=[\lambda_K]$$
This looks very fruitful to me (at least regarding our minus signs).
Any help would be greatly appreciated! Thanks in advance!
 A: Just to finish off the above argument:
Consider the following diagram:

Then we can continue in our above calculation
$$\begin{array}{rcl}
\partial_2(\left[ \mu_K^\star\right] \frown\left[ S^3\setminus F(B^2\times K)\right] )  &   =   &   -i^\star(\left[ \mu_K^\star\right] )\frown F_\star(-\left[ S^1\times K\right] )\\
                &   =   &   -F_\star(F^\star(i^\star(\left[ \mu_K^\star\right] ))\frown(-\left[ S^1\times K\right]) )\\
                &   =   &   -F_\star(\left[ S^1\times\left\lbrace \star\right\rbrace \right] ^\star\frown(-\left[ S^1\times K\right] ))\\
                &   =   & -(F_\star(-\left[ \left\lbrace \star\right\rbrace \times K\right] ))\\
                &   =   &   \left[ \lambda_K\right] = \left[ \partial F^\prime_K\right] = \partial\left[ F^\prime_K\right] .
\end{array}$$
where in the second to last row we used the explicit calcualtion of the cap product on a torus, this is:
Let $T=S^1\times S^1$ be the torus. Then the isomorphism
$$\frown\left[T\right]:H^1(T;\mathbb{Z})\rightarrow H_1(T;\mathbb{Z})$$
given by capping with the fundamental class satisfies
$$\left[\left\lbrace \star\right\rbrace \times S^1\right]^\star\mapsto-\left[S^1\times\left\lbrace\star\right\rbrace\right]$$
and
$$\left[S^1\times\left\lbrace\star\right\rbrace\right]^\star\mapsto\left[\left\lbrace\star\right\rbrace\times S^1\right].$$
