fundamental group of a graph

Question

let $G$ be a connected graph and $\Omega$ its universal covering. Let $\gamma_1,\dots,\gamma_r$ be free generators of $\Gamma:=\pi_1(G)$, $v\in\Omega$ be a vertex and $s_i$ a path from $v$ to $\gamma_i v$. Consider $T:=\cup_{\gamma} \gamma(\cup_{i=1}^r s_i)$, is it true that $\Omega$ is homotopy equivalent to $T$? If not, which conditions on $G$ ensure the equality? If $T$ homotopy equivalent to $\Omega$, is there an unicity statement like for every edge $e$ of $\Omega$ there exist unique $i$ and $\gamma\in\Gamma$ such that $e$ is homotopy equivalent to $\gamma s_i$?

References are also appreciate.

Thanks

Answer 1

There is an easy way to construct the universal cover of a connected graph. Essentially you pick any vertex of $G$, and for this vertex create a "central" vertex $r$ in $\Omega$. Then you recursively add for each vertex added in the previous iteration a vertex connected to that vertex in $\Omega$ for each "non-backtracking" vertex connected to that one in $G$. This means that if $G$ is a tree, then $\Omega = G$, and if $G$ has at least one cycle then $\Omega$ is a countably infinite tree which intuitively "separates" all non-backtracking paths into a tree, almost "uncurling", per se, all cycles.

For example, for the graph

We have the (beginning of) the universal cover:

Where the tree continues down infinitely in the same manner.

So to claim that $T$ is homotopy equivalent to $\Omega$ is to claim that it is contractible, or a tree itself. I believe that this is always true.