Question

Definitions: Let $G$ be a group wich acts without inversion on a connected nonempty graph $X$. We shall see that, if $X$ is a tree, then $G$ can be identified with the fundamental group of a certain graph of groups $(G,Y)$, where $Y=G\setminus X$. We begin with the construction of $(G,Y)$. Let $T$ be a maximnal tree of $Y$ and let $j:T\rightarrow X$ be a lifting of $T$. Let $A$ be an orientation of $Y$.

We shall extend $j$ to a section $j:edge Y\rightarrow edgeX$ such that $j \bar y= \overline{jy}$ ;

It suffices to define $jy$ for $y\in A-edgeT$ ; in this case we choose $jy$ so that $o(jy)\in vert jT$; we then have $o(jy)=jo(y)$. We also choose $\gamma_y\in G$ so that $t(jy)=\gamma_y*jt(y)$; this is possible because $t(jy)$ and $jt(y)$ have the same projection $t(y)$ in $Y$.

This is the original describtion of the book on side 54. So my question is the following:

1. If we choose an arbitrary $z\in edgeX$ with $jy=z$ and $o(jy)=o(z)\in vertjT$ it could be happen that $o(jy)\not= jo(y)$. But I think we should choose it such that this equation is satisfied, since $j$ should be extendend to a graph morphism, right? Or why is that clear?

2. Why does the $\gamma_y\in G$ exist? Okay. If $t(jy)$ and $jt(y)$ lying in the same equivalence class under the action of $G$, this is clear. But why does this two elements have to lie in the same equivalence class?

Answer 1

On 1.: $j$ is not meant to be a graph morphism. For example $Y$ could contain loops, which you have to 'break up' to 'lift' it to $X$.

On 2.: I think it is implicitly implied that you should choose $jy$ to have $y$ as projection in $Y$. If you assume this then they do lie in the same equivalence class.