Is there any way to describe the fundamental group of a connected graph by ping pong lemma? I already know that the fundamental group of a connected graph is the free group $F_n$, where $n$ is the number of its fundamental cycles. But I'm looking for a proof via ping-pong lemma. I was looking at the figure $8$, where its fundamental group is $F_2$, but no idea!


1 Answer 1


Here's a method of proof, which requires some setup. The rough idea is that the ping pong argument requires an action of the group, and to get an action of the fundamental group one uses the deck transformation action on the universal covering space.

Let's start with a few facts.

First, the 4-valent tree $T$, pictured in this answer of @ChristianBlatter, is a covering space of the figure 8.

Next, $T$ is simply connected (any tree is simply connected). Therefore $T$ is the universal covering space.

Now let's bring in the theorem that the fundamental group of the figure 8 acts as a group of deck transformations on the universal covering space $T$. And what's more, the proof gives a specific method that, given a loop, tells you which deck transformation of $T$ corresponds to the given loop.

So now I'm ready to set up the ping-pong proof.

In that figure of the universal covering space, let $t_1,t_2$ be the two deck transformations on $T$ that correspond to the two loops of the figure 8. Let $D \subset T$ be the fundamental domain for the deck action that contains the middle vertex and half of each of the four adjacent edges, so $D$ has four "endpoints": right, left, upper, and lower.

Now apply the above mentioned method for turning loops into deck transformations. It should be applied to the two loops in the figure 8 and their inverses, producing two deck transformations $t_1,t_2$ and their inverses. We get the following results:

  • $t_1$ moves $D$ to the right (shrinking $D$ apparently) so that $D \cap t_1(D)$ is the right endpoint of $D$. Furthermore, the entire component of $T-D$ to the right of the right endpoint is mapped rightward, hence into itself, by $t_1$.
  • $t_1^{-1}$ moves $D$ to the left so that $D \cap t_1^{-1}(D)$ is the left endpoint of $D$. Furthermore, the entire component of $T-D$ to the left of the left endpoint is mapped leftward, hence into itself, by $t_1^{-1}$.
  • $T_1$ moves $D$ up so that $D \cap T_2(D)$ is the upper endpoint of $D$. Furthermore, the entire component of $T-D$ above the upper endpoint is mapped upward, hence into itself, by $t_2$.
  • $T_1$ moves $D$ down so that $D \cap T_2^{-1}(D)$ is the lower endpoint of $D$. Furthermore, the entire component of $T-D$ below the lower endpoint is mapped downward, hence into itself, by $t_2^{-1}$.

And with this description, the ping-pong argument should be straightforward to fill in.

  • $\begingroup$ Another way to the result on graphs is to note that using the Van Kampen Theorem with a set $S$ of base points, one can prove that if $X$ is the union of open simply connected subsets then $\pi_1(X,S)$ is for suitable $S$ a free groupoid. Can this be related to the ping-pong argument? $\endgroup$ Jun 2, 2019 at 10:51

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