Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 19I / c

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 $\{b^i,b^{i}a\}$, $i=0,1,2$. 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.

The upper circle follows $e\to b\to b^2\to e$; the lower follows $a\to ab\to a{b}^2\to a$. The point paired with $b^i$ is $b^{i}a=a{b}^{-i}$, 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 $\ker \varphi$, identifying this kernel cover of a real projective plane wedged with a circle.

Yes, it retracts onto a graph. On each of the three spheres, choose an arc joining the two marked points. A sphere retracts onto this arc: identify it with the vertical meridian parameterized by the height coordinate, and send every point of a given height to the meridian point at that height. This map is continuous and fixes the meridian. Apply these three retractions and fix the lifted circles pointwise; the maps agree at all attachment points. The resulting connected graph has six vertices and nine edges, so its fundamental group is free of rank $\boxed{9-6+1=4}$.

No, it cannot deformation retract onto a graph. Each sphere contributes a degree-$2$ homology class independent of the other two. Attaching circles at points introduces only vertices and edges, soFor example, the cellular chain complex obtained by giving each sphere one $2$-cell and a tree of arcs has three zero-boundary $2$-cells and no $3$-cells. Every graph has zero degree-$2$ homology. A deformation retraction would induce a homotopy equivalence and an isomorphism on homology, which is impossible. An ordinary retraction has no such requirement.