The <Alexander numbering> of the three lobe regions is one and that of the central region is two, relative to the unbounded region numbered zero. Equivalently, abelianizing the three relators gives
$$
A-D+B=0,
\qquad B-D+C=0,
\qquad C-D+A=0.
$$
Thus $A=B=C$ and $D=2A$. The class $A$ is the positively oriented <meridian of a knot>, so
$$
\boxed{\phi(a)=\phi(b)=\phi(c)=\phi(\mu),
\qquad \phi(d)=2\phi(\mu).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 141 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-141/1/c/solution