= Solution
The kernel is a <normal subgroup> of index $6$, so the corresponding connected <covering space> is regular with six sheets labeled by $S_3$. Write $a=(12)$, $b=(123)$ and use right multiplication for path lifting. Above the <real projective plane>, right multiplication by $a$ pairs the labels, giving three copies of its two-sheeted <covering space> $S^2$. Each <sphere> has two marked antipodal points corresponding to its pair $\{b^i,b^ia\}$, $i=0,1,2$. Above the <circle>, right multiplication by $b$ gives two cycles of length three. Attach their six marked points to the identically labeled points of these three <spheres>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-2-cover.png]
{title=The six-sheeted cover: three spheres joined by two lifted circles}
The upper <circle> follows $e\to b\to b^2\to e$; the lower follows $a\to ab\to ab^2\to a$. The point paired with $b^i$ is $b^ia=ab^{-i}$, explaining the reversed lower labeling in the figure. On each summand these are genuine <covering maps>, and the attachment matches the six lifts of the wedge point. A small neighborhood of that point lifts to six disjoint copies of its neighborhood in the wedge. This verifies the <covering space> locally, and the action of $a,b$ is transitive, so it is connected. Its stabilizer is $\ker\phi$, identifying this <kernel cover of a real projective plane wedged with a circle>.
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