I'm assuming that the following question should be basically trivial, and that I'm just misunderstanding something basic, but some clarification would be much appreciated.
There is a section in my notes about the Nielsen-Schreier theorem that comes just after a large section on how we can find precisely one based covering space (up to equivalence) for each subgroup of the fundamental group of a path-connected simplicial complex. Here is the example given to us:
I know that, for a graph, generators of its (free) fundamental group are given exactly by those (oriented) edges that don't lie in some fixed maximal tree. I have two questions here. The first is trivial (and probably just a silly misunderstanding) and the second is slightly more confusing (for me):
- How do we understand that the bold edges in the diagram give a tree? It seems like it contains a loop (since the dotted edges join up).
- In the diagram there is just one edge $e$ not contained in the maximal tree $T$, so how come we have five generators for $H$? I know that this matches up with the number that we'd expect from the $nk-k+1$ formula (if we knew that $H$ was an index-four subgroup of $F_2$), but I don't see more than one edge in the diagram...
As some bonus questions (that I don't need to know the answers to for my exams, but intrigue me):
- How could we show that $H$ is an index-four subgroup in $F_2$? Or, could we show that $H$ is normal (is it even normal?) and then use the fact that the covering space corresponding to $H$ is the Cayley graph of $F_2/H$?
- This covering space looks a lot like a torus, but with just the vertices identified, not the sides. Is it in some way related?
I think you've misunderstood the description of $\tilde{X}$. It has four vertices (the corners of the square in the diagram) and eight edges: the four sides of the square and another four (dotted) edges joining pairs of adjacent vertices.
So it's a four-fold covering space of $X$, corresponding to the fact that the associated subgroup $H$ is index four in $F_2$.