So, I have seen fundamental polygons quite a few times now and I was always wondering what they are actually good for.
Let's take the sphere. It's fundamental polygon can be seen here image.
Does this mean, that if we identify the edge of this polygon in the way it is shown in this image which will give us some $X/R$ structure, where $X$ is the polygon and $R$ the respective equivalence relation that this structure is homeomorphic to the sphere?
If you identified the edges that you want to glue together, then you can go clockwise and look whether an arrow points in the direction you are walking or against you. Doing this you get $ABB^{-1}A^{-1} = AA^{-1}=Id$, so is it true then that we get the fundamental group by taking the number of different edges we have $A,B$ with the respective constraint? But what exactly is the constraint? Apparently in this case we would not have any, but if we have $AB^{-1}A^{-1}B$ for example. Why do we want this to be equal to one?
So why would the group in that case be given by $\langle A,B| AB^{-1}A^{-1}B=1 \rangle $?
Best Answer
The wikipedia page "Fundamental polygon", specifically the subsection entitled "group generators", has a serious mathematical error. You cannot derive a presentation for the fundamental group from the fundamental polygon using the side labels in the manner described on that page (and which you have copied), unless all of the vertices of the polygon are identified to the same point. In the picture you provided and which can be seen on that page, one opposite pair of vertices of the square is identified to one point on the sphere, the other opposite pair of vertices is identified to a different point on the sphere.
There is still a way to derive a presentation for the fundamental group from a fundamental polygon, but it is not the way described on the wikipedia page. In the sphere example of your question, you have to ignore one of the two letters $A,B$, keeping only the other letter. For example, ignoring $A$ and keeping $B$, you get a presentation $\langle B \mid B B^{-1} = 1 \rangle$, which is a presentation of the trivial group. The way you tell which to ignore and which to keep is by taking the quotient of the boundary of the polygon which is a graph with vertices and edges, choosing a maximal tree in that graph, ignoring all edge labels in the maximal tree, and keeping all edge labels not in the maximal tree.
On that wikipedia page, the Klein bottle and the torus examples are correct and you do not have to ignore any edge labels: all vertices are identified to a single point and the maximal tree is just a point. The sphere and the projective plane examples are incorrect: the four vertices are identified to two separate points, the maximal tree has one edge, and you have to ignore one edge label. The example of a hexagon fundamental domain for the torus is also incorrect: the six vertices are identified to two separate points, the maximal tree has one edge, and you have to ignore one edge label.