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I'm having trouble with a problem from Hatcher's "Algebraic Topology" (Section 1.3, Problem 33).

I'll provide a rough summary of the objects involved, but for a more precise description, have a look at examples 1.35 and 1.44 (pg 65 and 76 respectively) in Hatcher's book.

Here are the objects involved:

$X_{m,n}$ = Double mapping cylinder of the two maps $S^1 \to S^1$ defined by $z\mapsto z^m$ and $z\mapsto z^n$.

$T_{m,n}$ = A graph(in fact a tree) constructed as follows: Start with one vertex. Add m vertexes and an edge to the original vertex from each of the m vertexes. For each of those m vertexes, connect n-1 new vertexes, and for each of those, add m-1 new vertexes, and continue ad infinitum. The result is an infinite tree where each vertex has degree m or n. Call the root vertex $v_a$ and call one of its adjacent vertices $v_b$.

$T_{m,n} \times \mathbb{R}$ = The universal covering space of $X_{m,n}$.

$G_{m,n}$ = The group of deck transformations of the above covering map. It is generated by two elements $a$ and $b$ (since we have that $\pi_1(X_{m,n})=\langle a,b|a^m=b^n \rangle$). They act on $T_{m,n}$ by rotation about ${v_a}$ and ${v_b}$ respectively. They act on the $\mathbb{R}$ component by translations of length $1/m$ and $1/n$, giving a map $G_{m,n}\to \mathbb{Z}$, where (I think) $1$ is mapped to by elements which translates a length $1/{\mathrm{lcm}{(m,n)}}$. Let $K$ be the kernel of this map. There is a free action of $K$ on $T_{m,n}$.

Question: "Let $d=\mathrm{gcd}(m,n)$. Show that the graph $T_{m,n}/K$ consists of $m/d$ vertices labelled $a$, and $n/d$ vertices labelled $b$, together with $d$ edges joining each $a$ vertex to each $b$ vertex."

There is a second part to the problem, but I was able to solve it given this part.

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The space $X_{m,n}$, in Example 1.35, is not a double mapping cylinder. It can be written as a union of two mapping cylinders, but that is not the same thing. – Sam Nead May 11 '11 at 20:15
@Sam: These(pdf link) set of notes explicitly refer to it as a double mapping cylinder (page 23). Perhaps there are multiple meanings of the term? – alephzero314 May 11 '11 at 20:32
If you use the wikipedia definition with $X_1$=$X_2$=$S^1$, and with $f_1(z)=z^m$ and $f_2(z)=z^n$, you get exactly the space $X_{m,n}$. Sorry if I'm missing something obvious, but it looks identical to me. – alephzero314 May 11 '11 at 20:42
My mistake. Sorry. – Sam Nead May 11 '11 at 21:31
up vote 1 down vote accepted

Construct the space $X$ as follows. Let $I^2$ be a unit square, in $\mathbb{R}^2$. Glue the top and bottom of $I^2$ by translation to get a cylinder, $C$. That is, glue $(x,0)$ to $(x,1)$. To get $X$, glue the left side of $C$ by the map $(0,y) \mapsto (0, y + 1/m)$ and the right side via $(1,y) \mapsto (1,y+1/n)$. Note that the boundary of $I^2$ gives a one-skeleton to $X$. Pull this back to $C$: in $C$ there is one horizontal edge and there are $m$ vertical edges on the left and $n$ vertical edges on the right.

Now take the universal cover $U \cong T_{m,n} \times \mathbb{R}$; lift the one-skeleton given. Fix a vertex and say that it has height zero.

Exercise: Two vertices of the zero-skeleton $U^{(0)}$ have the same height and same label ($a$ or $b$) iff there is an element of $K$ taking one to the other.

Now, for any vertex $v$ of $T_{m,n}$ record the heights of all vertices of $U^{(0)}$ above $v$. It follows that there are $m/d$ classes of $a$ labelled vertices and $n/d$ classes of $b$ labelled vertices.

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Thank you very much. – alephzero314 May 12 '11 at 18:46

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