There are several threads about classification of all 3-sheeted covers of figure 8. For example, this one: How to classify 3-sheeted covering space for $S_{1}\vee S_{1}$?
It seems to me if tighten a few edges, we can deformation retract each of these into a wedge of circles of the same number, which results in a homotopy equivalence, as graphs.
But is it clear that they are also homotopy equivalent, as spaces? If I look at, for example, the table in Hatcher which gives a few presentations of the fundamental groups of 3-sheeted covers, it is not at all clear to me if they are isomorphic subgroups of $F_2$…
Thanks in advance!
Best Answer
It's a very special feature of $F_2$ that the isomorphism classes of its finite index subgroups are uniquely determined by their index. Geometrically, the point is that an $n$-sheeted cover of $S^1 \vee S^1$ has Euler characteristic $n$ times the Euler characteristic of $S^1 \vee S^1$, so $-n$, and a connected graph is uniquely determined up to homotopy equivalence by its Euler characteristic: every such graph is a wedge of $k$ circles for some $k$, which has Euler characteristic $1 - k$.
Hence every connected $n$-sheeted cover of $S^1 \vee S^1$ is homotopy equivalent to a wedge of $n+1$ circles, so has fundamental group $F_{n+1}$: algebraically, every subgroup of $F_2$ of index $n$ is abstractly isomorphic to $F_{n+1}$.