Since two days I'm thinking about the proof of Propostion 20, pages 44 and 45, in Serre's book Trees. My question is the following:
Why is f*p = id? The problem, i have, is that if I look at p(x), x an element of pi_1 (G, Y, P_0) some of the y_i in the path from P_0 to P_0 will be 1 under the action of p, because they are elements of T. So i don't understand why p should be injective!? And under the action of f, like it is constructed on page 44, these 1's won't be become back to the y_i's. I think if I look at a path y_1,...,y_n from P_0 to P_0. Then we have y_1,..,y_n-1 elements of T (the geodesic from P_0 to P_n-1 (P_n-1 unequal to P_0 and P_n = P_0). So all these y_i's (i element of 1,..., n-1) are 1 under the action of p? So how can p be injective, such that f could be the inverse of p?
Thanks for your help!
Edit by YC: on the OP's behalf, I am going to type out what Proposition 20 actually says. (I still don't think it requires much LaTeX competency, merely the effort of actually typing.)
Trees, Proposition 20. Let $(G,Y)$ be a graph of groups, let $P_0$ be a vertex of $Y$, and let $T$ be a maximal tree of $Y$. The canonical projection $p:F(G,Y) \to \pi_1(G,Y,T)$ induces an isomorphism of $\pi_1(G,Y,P_0)$ onto $\pi_1(G,Y,T)$.