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Hello!

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 $\pi_1(G,Y,T)$.

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Hi Eric, it might help those who don't have the book right in front of them if you write out the statement of the Proposition, your best rendering of it, as well as the portions of the book which you cite in your question. Also, LaTeX is supported on this site, just put the expression between dollar signs. –  j.c. Aug 19 '11 at 14:59
    
Hello! jc. Maybe I can scan the few pages of the book. And the I can send it to those, who are interested in the question and who could help me. If you speak German I also can give you al Link to a script in german, where the same thing ist proofed in the same way like it was done in the book. Thanks! –  Eric Aug 19 '11 at 15:27
    
Hello! In the folowing script the proof is the proof to "12.9 Bemerkung" p is defined in 12.5: math.tuwien.ac.at/~herfort/SEMINAR/SS09/grugra.pdf –  Eric Aug 19 '11 at 15:33
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I think rather than scan pages, you should attempt to write out those parts of the Proposition and the proof which you don't understand. This may also help you to localize your difficulties with it –  Yemon Choi Aug 19 '11 at 22:32
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Here are the 4 pages: Page 41: fotos-hochladen.net/uploads/page41ov4qgrs5kj.jpg Page 42 and 43: img4.fotos-hochladen.net/uploads/pages4243p28njv7lfw.jpg Page 44 and 45: fotos-hochladen.net/uploads/pages4445e9ix62uq51.jpg The Problem still ist the like i tried to explain it: Why should p be injective and why ist f*p = id. I think some of the y_i's in the second last line of the proof could be zero under p, since they are elements of T. So how can be be injective? –  Eric Aug 20 '11 at 9:24
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1 Answer

A loop $\gamma=y_1\cdot \ldots\cdot y_n$ of edges contained in $T$ is null-homotopic (because it's contained in the tree $T$), and so is also trivial in $\pi_1(G,Y,P_0)$.

Let's actually prove this. Because $T$ is a tree, it contains no non-trivial loops. Therefore, $\gamma$ contains a spur, ie a length-two subpath that crosses the same edge twice, once in either direction. In other words, $y_i=\bar{y}_{i-1}$ for some $i$. So we can cancel $y_i$ and $y_{i-1}$, which reduces the length of $\gamma$. We are now done by induction.

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