Universal cover of figure eight is simply connected

Question

I've been going through Viro's Topology book and I'm stuck at the universal cover of figure eight($ \mathbb{F}(a,b)$ is a free group generated by 2 elements).

I can imagine the universal cover of figure eight -- that is simply the Cayley graph of $F_2$ and it's clear that it is simply connected -- but I can't prove that strictly: i.e. I don't clearly understand how to construct a path between the points to shown the above task 35.P.. Furthermore I can't 'see' how to prove 35.Q..

Can anyone give me a hint?

Answer 1

A picture is worth a thousand words:

On the top left: the figure-eight space, with two generators $a, b$ for its fundamental group labeled. Top right: the universal cover $\Bbb{F}(a, b)$ with the lifts of $a, b$ labeled. Middle bottom: the same universal cover with some other elements of the fundamental group labeled. $a^{-1}$ and $b^{-1}$ represent going around the pictured loops in the opposite direction.

You can clearly see that in this picture, if I start at any element $x \in \Bbb{F}(a, b)$:

• the multiplication $xa$ corresponds to a step right from $x$ in the Cayley graph
• the multiplication $xb$ corresponds to a step up from $x$ in the Cayley graph
• the multiplication $xa^{-1}$ corresponds to a step left from $x$ in the Cayley graph
• the multiplication $xb^{-1}$ corresponds to a step down from $x$ in the Cayley graph

This should be very helpful as you think about how to demonstrate path-connectedness of the universal cover--if path-connectedness isn't built into your definition of a covering space of a path-connected space. The universal cover is simply connected by definition, I believe?

Answer 2

You seem comfortable thinking about this construction as the Cayley graph of $\mathbb{F}(a, b)$, the free group on $\{a, b\}$, so I will hint at arguments you can make in terms of that. (I will be slightly sloppy about distinguishing a graph as formal data vs its geometric realization as a topological space.)

Path-connected: Every Cayley graph is path-connected as a topological space. For a hint as to how to prove this rigorously, note that if a group $G$ has generating set $S$ then by definition every element $g$ can be written (possibly non-uniquely) as a composition $s_1^{\epsilon_1}s_2^{\epsilon_2}s_3^{\epsilon_3}\dots s_n^{\epsilon_n}$ for $s_i \in S$ and $\epsilon_i \in \{-1, 1\}$. For the vertex $v[g]$ corresponding to $g$ in a Cayley graph $\Gamma(G, S)$, there is a (topological) path from the vertex $v[e]$, the vertex corresponding to the identity element, to $v[g]$ through the vertices $v[s_1^{\epsilon_1}]$, $v[s_1^{\epsilon_1}s_2^{\epsilon_2}]$, $v[s_1^{\epsilon_1}s_2^{\epsilon_2}s_3^{\epsilon_3}]$, etc., where whenever you have $\epsilon_i = -1$ you have to travel backwards along an edge in the directed graph.

Aside: As a directed graph the Cayley graph is frequently not path connected. For example if $F$ is a free group on $S$ and $s\in S$ then there is no directed path from $v[e]$ to $v[s^{-1}]$ in the Cayley graph $\Gamma(F, S)$, even though there is one from $v[s^{-1}]$ to $v[e]$.

Simply-connected: It is a fact that an undirected graph is simply-connected iff it is a tree. Show that if $\Gamma(G, S)$ (considered as an undirected graph) is not a tree then $G$ has a non-trivial relation in terms of $S$, or in other words if $F$ is a free group on $S$ then $\Gamma(F, S)$ is a tree.

Answer 3

Let $T$ be the Cayley graph of $G=\mathbb F(a,b)$ with respect to $S=\{a,b\}$.

Let us first show that $T$ is path-connected.
Let $u,v\in G$ and let $w=u^{-1}v$. Suppose $w=s_1s_2\ldots s_n$ where the $s_i$'s are in $S$. Then we get a path from $u$ to $v$ whose vertices are $u$, $us_1$, $us_1s_2$, $\ldots$, $us_1s_2\ldots s_n=uu^{-1}v=v$.

The same argument shows that if $S$ is a generating set for a group $G$, then the Cayley graph of $G$ with respect to $S$ is connected. We now show that $T$ is simply-connected.

We define a sequence of trees $(T_n)_{n\in\mathbb N}$ as follows.

Let $T_0$ be the tree with a single vertex corresponding to the identity element $e$. Let $T_1$ be the tree with four edges whose leaves correspond to the words of length 1, namely $a,b,a^{-1},b^{-1}$. Let $T_2$ be the tree whose leaves correspond to the words of length 2. Let $T_n$ be the tree whose leaves correspond to the words of length $n$.

The interiors of the $T_n$'s give an increasing open cover of $T$. As the image of any loop in $T$ is a compact subset of $T$, it is contained in some $T_n$. As $T_n$ is a tree, it is a simply-connected subset of $T$.

Hence, it follows that $T$ is simply-connected.