I am learning covering spaces for the first time and came across the following problem:
Describe the covering space of the wedge of two circles that corresponds to the subgroup of $\pi_1(S^1\vee S^1) = \langle a, b\rangle$ generated by $\{a^2, b^2, (ab)^4\}$.
I realize that this is similar to question 12 on page 80 of Hatcher's Algebraic Topology, but not quite the same. There he asks for the covering space corresponding to the normal closure of $\langle a^b, b^2, (ab)^4\rangle$, not the subgroup group itself.
This is my attempt: Take the vertex of $S^1\vee S^1$ as its base-point, so that the paths traversing each of the two edges are the generators of the fundamental group. We then have to construct an oriented graph, with each edge labeled $a$ or $b$. Each vertex locally must look like the one vertex on $S^1\wedge S^1$: Two edges labeled $a$ (one pointing in and one pointing out) and two edges labeled $b$ with the same requirement.
The homotopy class corresponding to $a^2$ is a single loop, drawn bellow as one circle with two vertices and two edges going clockwise. Similarly, $b^2$ is a loop of two edges. For $(ab)^4$ we traverse only the upper half of the circle corresponding to $a^2$ and then six more edges, in total corresponding to $(ba)^3$. Finally, we traverse the upper half of the circle corresponding to $b^2$. Due to the local requirement for each vertex, we need additional edges added to most vertices, where we can glue an appropriate infinite tree with two generators where indicated by the gray disks.
Is this it? Did I get something wrong?