I am new to learning about finite rooted binary trees. This lemma below is from John Meiers book: Groups, Graphs and Trees. There is no aval proof in the book. I was just wondering is I could catch a few pointers on how I could solve this. Let $T_1$, $T_2$, and $T_3$ be finite rooted binary trees with the same number of leaves. Then $[T_3 \leftarrow T_2][T_2 \leftarrow T_1] = [T_3 \leftarrow T_1]$ and $[T_2 \leftarrow T_1]^{-1} = [T_1 \leftarrow T_2]$.
Finite Rooted Binary Trees
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trees
geometric-group-theory
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0Could you please explain what $[T_1\leftarrow T_2]$ means, for those of us who don't have access to the book? – 2012-05-19
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0Sorry about that. first dyadic divisions of [0,1] correspond with finite rooted bianry trees. $[T_1 \leftarrow T_2]$ is the thompson function where the domain is divided according to $T_2$ and the range according to $T_1$. A dyadic division of [0,1] is constructed by dividing [0,1] into [0, 1/2] and [1/2,1] and then one picks middles of the resulting pieces in a finite number of times. – 2012-05-19
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0Thanks. It would be best if you edit that information into the question itself. I'm not sure everyone would know what the Thompson function is, and you also haven't explained precisely how you obtain the diadic division of $[0,1]$ from a given rooted binary tree. Also, I don't understand the relevance of your last sentence, because you haven't used $\sim$ anywhere else. – 2012-05-19
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0@Jami You really have to define the function. – 2012-05-19