Fundamental group of the quotient of a polygon So I learned how to calculate the fundamental group of compact connected surfaces, this essentially comes from the fact that we know that they are the quotient of a polygon from a certain labelling scheme and we can use the seifert van-Kampen theorem to calculate its fundamental group.
However i have a question wich is, what happens if we try to do the quotient of a polygon with a certain labelling scheme, but we have more than one vertice in the quotient of the space. When we have only one we know that one of the open sets that we use has fundamental group $\mathbb{Z}*...*\mathbb{Z}$ $n$ times supposing we have $n$ different labels, but this uses the fact that they all go to the same vertice. What is going to be the fundamental group of that open set when we have more than over vertice in the quotient, is there a way of knowing it? Im guessing this question could be sum up in one question wich is, Can we calculate the fundamental group of the topological space obtained by a quotient of a polygon by any labelling scheme? Thanks in Advance.
 A: The method you learned for writing down a presentation of the fundamental group of a surface applies as well to your more general question. I'm not sure what generality of examples you've done, so what I write may be entirely familiar to you. 
To describe how this is done, consider a polygon $P$ and a labelling scheme under which each edge of $P$ is labelled with an orientation and a letter, and then all edges labelled the the same letter are identified to a single edge. Let $Q$ denote the quotient space. Notice: I am not assuming that the edges are identified in pairs: they could be identified in triplets; or higher multiplets; or perhaps some edge is a singlet, identified to nothing else whatsoever.
Under the quotient map from $P$ to $Q$, there is an induced identification of the vertices of $P$. The quotient of the boundary of $P$ is therefore a "graph" in $Q$ which I will denote $Q^{(1)}$, consisting of vertices and edges in which each endpoint of each edge is identified with some vertex. The whole of $Q$ is an example of a "2-dimensional complex", and $Q^{(1)}$ is its "1-skeleton" (eventually you'll run into this terminology when you discuss simplicial complexes or CW complexes).
The graph $Q^{(1)}$ is connected, because it is a quotient of the boundary of the polygon $P$ which is itself connected. In any connected graph there is a maximal tree, which is not usually unique. In the special case that $Q^{(1)}$ has just a single vertex, the maximal tree is just that vertex alone, but if $Q^{(1)}$ has more than one vertex then $Q^{(1)}$ will contain at least one edge. 
Choose a maximal tree $T \subset Q^{(1)}$. Let $e_1,...,e_n$ be the complete list of labels of oriented edges of $Q$ that are not in $T$. These will edges are be in one-to-one correspondence with the generators of $\pi_1(Q)$, which I will denote $g_1,...,g_n$. 
It is possible that $T=Q^{(1)}$ (this happens, for example if you have a square with boundary labelled $abb^{-1}a^{-1}$, and so $Q$ is homomorphic to the 2-sphere). If that is so then there are no generators, the presentation is trivial, and the fundamental group is trivial, and you're done.
Assuming that $T$ is not equal to $Q^{(1)}$, and so the list of generators $g_1,...,g_n$ is not empty, the presentation will also have one relator $R$ (this part is probably familiar to you). 
You're going to write down the letters of $R$ one at a time, each letter having the form $g_i$ or $g_i^{-1}$ for some $i=1,...,n$. To write down $R$, you start at a vertex of the polygon $P$ and you walk around the boundary of $P$ until returning to where you started. As you walk along, you look at each edge that you cross and you ask yourself: under the quotient map to $Q$, is this edge in the maximal tree $T$ or is it not in $T$? 
If that edge is not in $T$, then it is one of $e_1,...,e_n$, and you write the next letter of the relation $R$ to be the corresponding generator $g_1,...,g_n$ with a $+1$ exponent if you are crossing the edge in the positive direction, and a $-1$ exponent if you are crossing the edge in the negative direction. 
But if that edge is in $T$ you skip it and you write down nothing. 
When you have returned to the vertex, you have written down the relator $R$, and then you have a presentation
$$\pi_1(X) = \langle g_1,...,g_n \mid R \rangle
$$
Since you have seen the calculation of the fundamental group of a compact connected surface, you should be able to recognize that this is a generalization of that calculation. Also, presumably you know how to use the Seifert-Van Kampen Theorem for the case of the surface, and the exact same proof applies to justify this calculation.
