Let \(a\) and \(b\) be the generators of \(\pi_1(S^1 \vee S^1)\) corresponding to the two \(S^1\) summands. Draw a picture of the covering space of \(S^1 \vee S^1\) corresponding to the normal subgroup generated by \(a^2\), \(b^2\), and \((ab)^4\), and prove that this covering space is indeed the correct one.
Solution
Let \[ F=\pi_1(S^1\vee S^1)=F(a,b) \] and let \[ N=\langle\!\langle a^2,b^2,(ab)^4\rangle\!\rangle \] be the indicated normal closure.
1. The quotient group is\[ F/N\cong \langle a,b\mid a^2=b^2=(ab)^4=1\rangle \cong D_8, \]the dihedral group of order \(8\).
Put \[ r=ab. \] Then \(r^4=1\), and since \(a^2=b^2=1\) one has \[ a r a =a(ab)a =ba =r^{-1}. \] Thus the presentation becomes \[ \langle r,a\mid r^4=a^2=1,\ ara=r^{-1}\rangle, \] which is the standard presentation of the symmetry group of a square. Every element has one of the eight forms \[ 1,r,r^2,r^3,a,ar,ar^2,ar^3, \] and these are realized by distinct square symmetries, so the group has exactly eight elements.
2. Construct a covering graph \(\widetilde X\) with one vertex \(v_g\) for each\[ g\in D_8. \]For every \(g\) attach one oriented \(a\)-edge\[ v_g\longrightarrow v_{ga} \]and one oriented \(b\)-edge\[ v_g\longrightarrow v_{gb}. \]Map every \(a\)-edge homeomorphically around the \(a\)-circle of \(S^1\vee S^1\) and every \(b\)-edge around the \(b\)-circle.
At each vertex there is exactly one outgoing and one incoming edge of each label. Hence a small star neighborhood of the wedge point lifts to a disjoint union of star neighborhoods of the vertices \(v_g\), each mapped homeomorphically to the base star. Away from the wedge point each edge maps locally homeomorphically to the corresponding circle. Thus \[ p:\widetilde X\to S^1\vee S^1 \] is an eight-sheeted covering.
3. A picture of \(\widetilde X\) is obtained by arranging the eight vertices in the cyclic order\[ 1,\ a,\ ab,\ aba,\ (ab)^2,\ a(ab)^2,\ (ab)^3,\ a(ab)^3 \]and joining consecutive vertices alternately by \(a\)- and \(b\)-labelled edge pairs.
Right multiplication by \(a\) and \(b\) moves between the two cosets in exactly this alternating pattern. Because \(a^2=b^2=1\) in the quotient, the two directed lifts of a given labelled base loop between a pair of adjacent vertices are distinct parallel edges with opposite endpoint order. After eight alternating steps one returns to the initial vertex because \[ (ab)^4=1. \] This is the labelled eight-vertex Cayley covering graph of \(D_8\) with generators \(a,b\).
4. The subgroup \(K=p_*\pi_1(\widetilde X,v_1)\) contains \(N\).
The monodromy homomorphism of the labelled cover is the quotient map \[ \rho:F(a,b)\to D_8 \] given by right multiplication on the vertex labels. A word represents an element of \(K\) exactly when its lift from \(v_1\) is closed, equivalently when \(\rho\) sends it to the identity. Now \[ \rho(a^2)=\rho(b^2)=\rho((ab)^4)=1. \] Therefore the kernel of \(\rho\), hence \(K\), contains the normal closure of these three elements. Thus \[ N\subseteq K. \]
5. Both \(N\) and \(K\) have index \(8\) in \(F\).
By 1, \[ [F:N]=|F/N|=8. \] By 2, the covering has eight sheets, so the covering-subgroup index theorem gives \[ [F:K]=8. \]
6. Therefore\[ K=N. \]
The inclusion \(N\subseteq K\) from 4 together with equality of finite indices from 5 forces equality.
7. Hence the eight-sheeted labelled Cayley graph in <1>2–<1>3 is exactly the covering corresponding to the normal subgroup generated by\[ a^2,\qquad b^2,\qquad (ab)^4. \]
This is 6 and the classification of connected based coverings by subgroups of the fundamental group.