Prove that $F_2$ is a subgroup of $F_3$, and construct a covering space of $S^1\vee S^1\vee S^1$ corresponding to this subgroup

Question

As the title explains, I'm working on a question that asks me to prove that $F_2$ is a subgroup of $F_3$, and construct a covering space of $S^1\vee S^1\vee S^1$ corresponding to this subgroup.

If take $F_3$ = $\langle a, b, c\rangle$ then we can just take the subgroup generated by $\langle ab, c\rangle$ and it's easy to find an isomorphism from this to $F_2$ by mapping the generators to each other, but I'm not sure how I'd go about finding a covering space that corresponds to it.

Any suggestions would be appreciated.

Answer 1

Let $G$ be the base graph, a rose with three petals labelled $a$, $b$, $c$.

To build the covering space, start with two loops: one subdivided into two edges labelled $a$ and $b$, and the other loop left unsubdivided and labelled $c$. That's the core of your covering space, and I'll denote it $X_1$. It has two vertices, $v_1$ of valence $2$ and $v_2$ of valence $4$. Let $f_1 : X_1 \to G$ be the map which takes the $a$ edge of $X_1$ to the $a$ edge of $G$, and the same for $b$ and $c$ edges. That map $f_1$ is the beginning of the covering map.

Now I'll describe an inductive construction, defining graphs $$X_1 \subset X_2 \subset X_3 \subset \cdots $$ and maps $f_i : X_i \to G$ such that $f_{i-1}=f_i \mid X_{i-1}$ for all $i$. The ultimate covering space will be $X = X_1 \cup X_2 \cup X_3 \cup \cdots$, and the ultimate covering map will be $f : X \to G$ whose restriction to $X_i$ is $f_i$. So, of course, $X$ will be an infinite covering space.

I'll describe the first step of the induction. Look at the two vertices of $X_1$. Neither of them is "complete", meaning that neither has valence $6$ with $6$ incident directions labelled $a,a^{-1},b,b^{-1},c,c^{-1}$ (meaning: one $a$ pointing towards, one $a$ pointing away, and similarly for $b$ and $c$). Take the lowest incomplete one, which is $v_1$. The two incident directions to $v_1$ are labelled $a^{-1}$ and $b$. So, add four more edges, attaching an endpoint of each to $v_1$, labelling one of them $a$ pointing away from $v_1$, another $v$ pointing towards $v_1$, and the third and fourth labelled $c$, one pointing towards and one away from $v_1$. Let $X_1$ union those four edges by $X_2$. The vertex $v_1$ is not complete, but we've added four more incomplete vertices, call them $v_3,v_4,v_5,v_6$.

Perhaps you can now see the induction step: in $X_i$, take the lowest incomplete vertex (which, I think, will always be $v_i$), and add edges to complete it, forming $X_{i+1}$.