= Solution
Orient the diagram and assign its regions an <Alexander numbering> with $R_0$ numbered zero. The <abelianization> $\phi:\pi_1(E_K)\to\langle t\rangle$ sends $a_i$ to $t^{m_i}$, where $m_i$ is the number of $R_i$. Since $R_1$ is adjacent to $R_0$, $m_1=\pm1$.
Form the square matrix
$$
A_1=\left(\phi\!\left(\frac{\partial w_j}{\partial a_i}\right)\right)_{
1\leq j\leq n-1,\ 2\leq i\leq n}
$$
from the <Fox calculus>[Fox derivatives] with respect to $a_i$ for $i>1$. This is the <Alexander matrix> with the $a_1$ column deleted. The Fox identity implies that its maximal minors differ by the factors $\phi(a_i)-1$, and the standard presentation of the <Alexander module> therefore gives
$$
\det A_1\doteq\frac{t^{m_1}-1}{t-1}\Delta_K(t)\doteq\Delta_K(t),
$$
because $m_1=\pm1$. Thus $\Delta_K(t)$ is $\det A_1$ up to a unit $\pm t^r$.
Back to article page
