# Universal cover or Bass-Serre tree: difference between definitions given by Bass and Serre

Let $$(\mathbb G,\Gamma)$$ be a graph of groups. A $$G$$-path from $$u_0$$ to $$u$$ is $$g_0e_1g_1\cdots e_{n}g_n,$$ where $$e_1\cdots e_{n}$$ is a walk in $$\Gamma$$ from $$u_0$$ to $$u$$ and each $$g_i\in G_{s(e_i)}$$. We denote the set of all $$\mathbb G$$-paths from $$u_0$$ to $$u$$ by $$\pi[u_0,u]$$. Let $$\mathbb F(\mathbb G,\Gamma)$$ be the group generated by the vertex groups and the elements $$e$$ of edge $$\Gamma$$, subject to the relations $$\{s_e(g)e=et_e(g)\mid g\in G_e,e\in E(\Gamma)\},\{e^{-1}=\bar{e}\mid e\in E(\Gamma)\}$$

Recall that the fundamental group of $$(\mathbb G,\Gamma)$$ on the base point $$u_0$$ is the set of all elements $$g_0e_1g_1\cdots e_{n}g_n$$ in $$\mathbb F(\mathbb G,\Gamma)$$, where $$e_1\cdots e_{n}$$ is a closed walk from $$u_0$$ to $$u_0$$. In other words, the fundamental group of $$(\mathbb G,\Gamma)$$ is $$\pi[u_0,u_0]$$.

My question is regarding the universal cover (Bass-Serre tree) of $$(\mathbb G,\Gamma)$$.

In the book "Trees'' by Serre page 51, the universal cover is defined in the following way:

The vertices are the left cosets of the vertex groups in $$\pi[u_0,u_0]$$, i.e.

$$\bigcup_{u\in V(\Gamma)} \pi[u_0,u_0]G_u$$

However in a paper by Bass "Covering theory for graphs of groups ", he defined in the following way:

The vertices are $$\bigcup_{u\in V(\Gamma)} \pi[u_0,u]G_u$$

My question is that

Are they $$\pi[u_0,u_0]$$-isomorphic?or they are completely different?

If yes, I really wonder to know the $$\pi[u_0,u_0]$$-isomorphism map.

To address a small point: the sets $$\pi[u_0,u]$$ should really be considered as their images in $$\mathbb F(\mathbb G,\Gamma)$$. This has the effect of doing some algebraic version of passing to homotopy classes of paths rel endpoints. For example, if $$e$$ is an edge of $$\Gamma$$, then $$e\bar e e$$ and $$e$$ should be considered as equivalent. This is already what happens when constructing the universal cover of a graph, say.
Modulo agreement on that point, my claim is that the definitions of the vertex sets of the universal cover of $$(\mathbb G,\Gamma)$$ are equivalent. Here is how to see it. Fix a choice of spanning tree $$T$$ in $$\Gamma$$. For every vertex $$u$$ of $$\Gamma$$, there is a unique path $$\sigma_u$$ from $$u_0$$ to $$u$$ that stays entirely within $$T$$. We can think of $$\sigma_u$$ as an element of $$\pi[u_0,u]$$ by sending $$\sigma_u$$ to (the equivalence class of) the ordered sequence of edges it traverses.
We get maps $$\pi[u_0,u_0] \to \pi[u_0,u]$$ and $$\pi[u_0,u]\to \pi[u_0,u_0]$$. The former sends a class $$\tau \in \pi[u_0,u_0]$$ to the class of the concatenation $$\tau\sigma_u$$. The latter sends $$\rho \in \pi[u_0,u]$$ to the class of the concatenation $$\rho\bar\sigma_u$$. Since the classes $$\tau$$ and $$\tau\sigma_u\bar\sigma_u$$ are equal, it’s easy to see that these maps are inverse bijections.
• Don't you mean that for every vertex $u$ there is a unique path that says entirely within $T$ (not $\Gamma$)? – Max Horn Dec 30 '19 at 8:44