Covering space of $S^1\vee S^1$

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Problem 12 Let $a$ and $b$ be the generators of $\pi_1(S^1 \vee S^1)$ corresponding to the two $S^1$ summands. Draw the covering space of $S^1 \vee S^1$ corresponding to the normal subgroup generated by $a^2, b^2,(a b)^4$.

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The covering space is a graph with 8 vertices labeled as the elements of the group $G = \langle a, b \mid a^2 = 1, b^2 = 1, (ab)^4 = 1 \rangle$: $e$, $a$, $b$, $ab$, $ba$, $aba$, $bab$, $abab$.

The edges are as follows, with each connection consisting of two parallel edges:

• Lifts of the first $S^1$ (labeled $a$): $e$ to $a$, $ab$ to $aba$, $abab$ to $bab$, $b$ to $ba$.
• Lifts of the second $S^1$ (labeled $b$): $e$ to $b$, $a$ to $ab$, $ba$ to $bab$, $aba$ to $abab$.

This graph can also be visualized as an 8-cycle\[\def\lr#1{\overset{#1}\leftrightarrows}e \lr{b} b \lr{a} ba \lr{b} bab \lr{a} abab \lr{b} aba \lr{a} ab \lr{b} a \lr{a} e\]
To prove this is the correct covering space, note that it is constructed as the Cayley graph of $G$ with generating set $\{a, b\}$, accounting for $a^{-1} = a$ and $b^{-1} = b$ by doubling the edges to preserve the local structure of the base space (degree 4 at the vertex). Since $G \cong \pi_1(S^1 \vee S^1)/N$ where $N$ is the normal subgroup generated by $a^2$, $b^2$, and $(ab)^4$, this is the regular covering space with deck transformation group $G$. The degree is 8, matching $|G| = 8$, and the Euler characteristic is $8 - 16 = -8$, consistent with an 8-sheeted covering of a space with Euler characteristic $-1$.