Detail in Massey's Classification Theorem for Compact Surfaces

Question

So, I am going through Massey's Algebraic Topology: an Introduction, and I got stuck in a little detail in the proof of Theorem 5.1 of Chapter one (the Classification Theorem for Compact Surfaces).

In the very end of the fourth step, the author proves that edges of "the first kind" must appear separated by another pair of this kind (simbolically, it looks something like $c\dots d \dots c^{-1}\dots d^{-1}$). And then they argue the following:

To prove this assertion, assume that the edges labeled с are not separated by any other pair of the first kind. Then our polygon has the appearance indicated in Figure 1.20. Here A and В each designate a whole sequence of edges. The important point is that any edge in A must be identified with another edge in A, and similarly for B. No edge in A is to be identified with an edge in B. But this contradicts the fact that the initial and final vertices of either edge labeled "c" are to be identified, in view of step number three.

Kinsey elaborates a little more on the argument in Topology of Surfaces, saying that if we had only pairs of the "second kind" between the edges labeled $c$ we would have at least two different vertices on the polygon, contradicting the previous step of eliminating all but one vertex. And then, they conclude that the only possibility that doesn't violate that last condition is to have pairs of "the first kind" the way I said above.

However, I can't even understand why that would imply that the vertices are different... Can someone help me in this argument?

Answer 1

Your language lost some precision when you paraphrased from Kinsey regarding vertex identifications. So let me rewrite the implication you are asking about, using the more precise language of vertex identification as in Massey:

If each edge in $A$ is identified with some other edge in $A$ then for any vertices $v,w$ such that $v$ is identified with $w$, $v$ is in $A$ then $w$ is in $A$. Therefore, no vertex in $A$ is identified with a vertex in $B$.

As for why this statement is true, we'll use the fact that two different vertices $v \ne w$ of the polygon are identified if and only if they are forced to be identified by some sequence of edge identifications. More precisely, by definition $v$ and $w$ are identified if and only if there exists a chain of vertices $$(*) \qquad v=v_0 \ne v_1 \ne ... \ne v_n=w $$ such that for each $i$, the two vertices $v_{i-1}$ and $v_i$ are identified to each other by a single edge identification.

So, assuming that $v$ is in $A$, we prove that $w$ is in $A$ by induction on the length $n$ of the chain of vertices $(*)$.

For the base case of the induction where $n=1$, we just have to check on edge identification: if $v \ne w$ are two vertices of the polygon, if $v$ is an end point of one edge labelled $e$ (which I'll call "the first $e$ edge"), and if $w$ is the corresponding endpoint of the other $e$ edge to which $v$ is identified (which I'll call "the second $e$ edge"), and if $v$ is a vertex in $A$, then $w$ is also a vertex in $A$. There are two cases to consider:

• The label $e$ is different from $c$. Because $v$ is in $A$, it follows that the first $e$ edge is in $A$. Because any edge in $A$ is identified with another edge in $A$, it follows that the second $e$ edge is also in $A$. Therefore $w$ is in $A$.
• The label $e$ is the same as $c$. So the first $e$ edge is one of the two $c$ edges, and $v$ is its final vertex. It follows that the second $e$ edge is the other $c$ edge, and $w$ is its final vertex. Therefore $w$ is in $A$.

For the inductive step, assume that statement $(*)$ is true for any chain of length less than $n$. For the chain of length $n$ as shown in $(*)$, we break it into the length $n-1$ chain $v=v_0,...,v_{n-1}$ followed by the length $1$ chain $v_{n-1},v_n=w$. Applying the inductive step to that length $n-1$ chain it follows that $v_{n-1}$ is in $B$. Then applying the inductive step to that length $1$ chain it follows that $w$ is in $B$.