Generating set of the image of fundamental group of a covering space I am doing problem 1 in this exam.
I am really confused about the question about generating set. What I am asked for? For (a), I know the rank should be 4 and the index should be 3. I also know we can consider the maximal tree that consists of $x_1$ and b,-b as edges to find the fundamental group of covering space. But what is the generating set the question asks for? For (b), if we choose a different base, what will happen? (I find I know this, the $H_1$ and $H_2$ should be conjugate)
 A: To construct a generating set on a graph we take advantage of the fact that a graph is homotopy equivalent to the wedge of circles obtained by contracting its maximal tree. In your example, if you contract the maximal tree you are left with four circles wedged together. You need to determine which sequence of edges in the original graph are homotopy equivalent to the circles you end up with. For example, after contracting the maximal tree the loop $a$ at $x_0$ is one such generator, and the loop $bab$ contracts to a circle in the wedge. Does this give you the idea?
If you change the basepoint in the cover you create a conjugate subgroup, but when you write out the generators for $H_1$ and $H_2$ you should be able to guess the conjugate element. Of course I'm sure your textbook gives an explanation of the change of basepoint in a covering space.
A: First, start by writing out a generating set for the fundamental group of the covering graph $\tilde{X}$. This is easily done by quotienting the maximal tree, as $\tilde{X}$ and $\tilde{X}/T$ are homotopy equivalent since $T$ is contractible, and $\tilde{X}/T$ is a wedge of circles. To be explicit, the generator set for $\pi_1(\tilde{X}/T)$ consisting of classes of loops going around a circle in the wedge can be used to obtain a generator set for $\pi_1(\tilde{X})$ by the isomorphism $\pi_1(\tilde{X}) \cong \pi_1(\tilde{X}/T)$.
Each circle in the wedge in $\tilde{X}/T$ comes from some path in $\tilde{X}$. For example, if you contract the maximal tree $T$ which has one vertex $x_1$, the edge $b$ after it, the node it ends at, then again the edge $b$ after it plus the node it ends at (so, $T$ is has 3 vertices and two edges, each between consecutive pairs of them), then you get a generator circle in $\tilde{X}/T$ given by image of the next $b$ edge by the quotient map $\tilde{X} \to \tilde{X}/T$. Thus, the loop $bbb$ is a generator of $\pi_1(\tilde{X})$ coming from the standard generating set for $\pi_1(\tilde{X}/T)$.
You have to determine a generating set for $p_*\pi_1(\tilde{X})$. Thus, it's enough to see the markings of the paths in $\tilde{X}$ which maps to the standard generators of $\tilde{X}/T$ under the quotient map $\tilde{X} \to \tilde{X}/T$ (e.g., $b^3$ would be a generator of the image in this way).
