I have a specific question about a part in the proof of this theorem:
Let $S$ be a compact surface, then triangulate $S$ with $\phi:T \rightarrow T'$(where $T'$ are triangles in the plane) and represent the identification of sides of the triangles $T'$ as an identified space homeomorphic to closed disk. Take the boundary of the identified space and represent it as a sequence of segments to be identified (ex. $aba^{-1}bcde...e^{-1}...d...e^{-1})$ where $k^{-1}$ represents an identification in the opposite direction of $k$. Classify "first kind segments" and "second kind segments" as those segments that appear in a boundary representation as "$...a....a...$" and "$...a...a^{-1}...$", respectively.
Then eliminate all adjacent edges of the first kind and transform the shape into a polygon where all vertices are identified to a single point. Additionally, transform the shape such that all remaining "second kind" edges become adjacent.
Here's where I don't understand:
Assume that there is at least one pair of edges of the first kind. Then we assert that there is at least one other pair of edges of the first kind such that these two pairs separate one another. To prove this assertion, assume that there is only one "first kind" pair of edges separated by $A$ and $B$ (Two separated sequences of the form $a_1a_1a_2a_2...a_na_n$ and $b_1b_1...b_mb_m$, respectively). The important point here is that any edge in $A$ must be identified with another edge in $A$, and similarly for $B$. But this contradicts the fact that the initial and final vertices of either of the "first kind" edges are to be identified, in view of the fact that all vertices are to be identified to a single vertex.
Where is the contradiction here? Because if there are two separate "first kind" edges then isn't this still a problem?