# Are 3-sheeted covers of figure 8 homotopy equivalent to each other?

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 $$F_2$$...

• It's not clear to me what you are asking. Two graphs which are homotopy equivalent as graphs are also homotopy equivalent as spaces, because a graph is a special kind of space. On the other hand, you seem to be overlooking a very important point: in that link, and in Hatcher's book, one is classifying covering spaces up to covering equivalence, not up to homotopy equivalence. – Lee Mosher Dec 3 '18 at 22:11
• I am just confused... On one hand, it seems that I can collapse edges of those covering spaces. If say I take all the 3-sheeted covers of figure 8, and collapse one edge between each pairs of vertices, I get a wedge of circles. I think I can get a wedge of the same number of circles for all the 3-sheeted covers, which tells me that all the 3-sheeted covers are homotopy equivalent. On the other hand, by looking at the presentations of their fundamental groups, I really couldn't tell if they have isomorphic fundamental groups... which in turn makes me doubt if they are really homotopy equivalent – chikurin Dec 3 '18 at 22:32
• Or do we actually can see the groups are isomorphic to each other by manipulating those relations? – chikurin Dec 3 '18 at 22:33
• Alright, then I understand your question to be not at all about covering equivalence, instead just about homotopy equivalence. I can given an answer but I don't have time for now. – Lee Mosher Dec 3 '18 at 22:41

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 $$F_{n+1}$$.
• Thanks! Should this work for an arbitrary connected (finite?) graph, whose $n$-sheeted covers are also connected (finite) graphs each of which can be deformed into wedge of a few circles? This seems even harder for me to believe... But probably I just need to believe it... – chikurin Dec 3 '18 at 23:37
• @chikurin: yes. The conclusion is that every subgroup of $F_{m+1}$ of index $n$ is abstractly isomorphic to $F_{mn+1}$. I agree that this is quite surprising stated as a bald fact about groups. – Qiaochu Yuan Dec 3 '18 at 23:38