# How do I find the number of vertices on a planar diagram of a surface?

Cheers,

I have a question which I just do not seem to see the answer for:

I am proving the classification theorem for compact surfaces and use planar diagrams as representation of the surfaces. I follow the combinatorical proof of Seifert and Threlfall with pasting and cutting on the planar diagrams. One of its steps says that the planar model gets reduced down to one single vertex.

Now, if I only have an algebraic representation of a planar model, e.g. bbcca^(-1)dda how do I know how many vertices there are? So far I think that if a vertex is the terminal point of an edge z, and the initial point of an ege y for example, then this must be ahered throught the whole planar diagram.

In this exact example above why cant it all be identified to one vertex? Or more specific, why cant any vertex out of bbcc be identified with one out of dd?

I am thankful for any help!

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I find it easier to first convert the algebraic representation into a polygon with edges identified in pairs. But let me write it out with capital letters for the vertices: $$AbBbCcDcEa^{-1}FdGdHaA$$ Now vertex $A$ is at the trailing end of an edge labeled $b$, and so is vertex $B$, so they are the same; $$1b1bCcDcEa^{-1}FdGdHa1$$ $1$ is at the front end of an edge labeled $b$, and so is $C$, so they are equal: $$1b1b1cDcEa^{-1}FdGdHa1$$ Now $1$ and $D$ are at the trailing edge of $c$, so they are equal: $$1b1b1c1cEa^{-1}FdGdHa1$$ Now $1$ and $E$ are at the front end of an edge labeled $c$, so they are equal: $$1b1b1c1c1a^{-1}FdGdHa1$$ Now $F$ and $H$ are in the same relation to $a$, so they are equal to each other: $$1b1b1c1c1a^{-1}2dGd2a1$$ Finally, $2$ and $G$ are in the same relation to $d$, so they are equal: $$1b1b1c1c1a^{-1}2d2d2a1$$ So we conclude that there are two vertices.