The universal cover of the figure-8 is the Cayley graph of the free group on $2$ generators with generating set $\{a,b\}$. So it is a tree. I know that finite trees are contractible. But this Cayley graph has too many edges and I don't know if I can contract this in finite time.
So my question is whether this universal cover contractible? If no why?
Thanks in advance!
Any non-empty locally finite tree is contractible. To see this, let $(T,d)$ be such a tree realized as a metric space, i.e. any edge has length one. Fix some $x_0 \in T$. For any $y \in T$, let $c_y : [0,1]\to T$ be the unique path of length 1 and constant speed connecting $x_0$ to $y$, i.e. $c_y(0)=x_0$ and $c_y(1) = y$ as well as $d(x_0, c_y(t)) = t \cdot d(x_0,y)$. Now define the following homotopy: $$ H : T\times [0,1] \to T; \quad (y,t) \mapsto c_y(t).$$
It is continuous by uniqueness of $c_y$, and clearly $H(\cdot,1) = \operatorname{id}_T$ and $H(\cdot,0) \equiv x_0$, hence $T$ is contractible.