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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?

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backslash overline curly brackets gives you a bar over more than one character. I edited it. – Stefan Geschke Sep 13 '11 at 9:43
thanks, Stefan. – Eric Sep 13 '11 at 9:56

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.

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Why couldn't be $j$ a graph morphism? Why do I have to break up these loops you mentioned? Couldn't they be mapped on something in $X$? And if j is not a graph morphism, is it implicity implied, too, that $o(jy)=jo(y)$? Then I would say, that the construction is a little bit impresicely, because it is concluded, that if we only choose $o(jy)\in vertjT$ all the other facts are immediately clear. – Eric Sep 13 '11 at 13:08
Let me clarify what is going on with an example: Assume that we are dealing with the case where the graph $Y$ is a 4-cycle. A spanning tree $T$ of this graph is then a path on 4 vertices. One can lift this spanning tree $T$ into the tree $X$ easily. However the question is now what to do with the remaining edge $y$ in $Y$. AS $X$ is a tree, and because of the way we embed $T$ in it, it is impossible to choose a lifting of $y$ in $X$ such that we obtain a graph morphism. (the remaining part is for the next comment) – Koen S Sep 14 '11 at 9:32
What Serre does is to pick an orientation on the remaining edge (the $A$), so we are only trying to lift $y$ but not $\overline{y}$. The lift of $y$ is then chosen such that it attaches to the embedding of $T$ in $X$ in the lift of the point $o(y)$ (this is what is meant by "in this case we choose $jy$ so that $o(jy) \in vertjT$; we then have $o(jy)=jo(y).$"). The resulting lift (or better said section) of $Y$ in $X$ is then a path on 5 vertices. What I said was implicit might be contained in the definition of a "section", which I cannot find in the book right away. I hope this helps. – Koen S Sep 14 '11 at 9:43
Thanks. Hmm. Okay. $X$ doesn't have to be a tree for the construction, right? At first it is only an connected nonempty graph. Only in the following Theorem in the book the situation where $X$ is a tree is involved. But you are still right, because it is possible, that a fundamental domain doesn't exist. Right? Another question is: If we take a lift $j$ of $T$ in $X$. Is it true, that the image of $t\in vertT$ lies in the equivalence class of $t$, such that $G∗(jt)=t$? Or is this assumption wrong?And why do we find this $\gamma_y\in G$? How would you explain it in your example? Thank you. – Eric Sep 15 '11 at 10:27

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