Fundamental group of the $S^2$ with the labeling scheme I know that the labeling scheme of $S^2$ is $aa^{-1}$ which should imply the following representation of the fundamental group:
$$\langle a | aa^{-1}=1\rangle$$
but this is not trivial unlike the fundamental group. How do I square this?
 A: Note that for a surface $X$ given by a given labeling scheme $a_1^{\epsilon_1}\cdots a_n^{\epsilon_n}$ where each $\epsilon_i = \pm 1$ (not necessarily all presented in order; e.g., $a_1 a_2 a_1^{-1} a_2^{-1}$ is allowed), this is equivalent to obtaining $X$ via the following construction.
Take a wedge of $n$ circles labeled $a_1, \ldots, a_n$, and call this space $X^1$. Then attach $D^2$ to $X^1$ via the map $S^1 \to X^1$ defined as follows: Divide the domain $S^1$ into a number of parts equal to the number of symbols in the labeling scheme, and then for each chunk of the domain $S^1$ we trace the named circle in $X^1$ either forwards or backwards according to the labeling scheme. Call the result $X$.
This wording is complicated, so here are some examples:
For the labeling scheme $a_1 a_2 a_1^{-1} a_2^{-1}$, $X^1$ is a wedge of $2$ circles labeled $a_1$ and $a_2$. We think of the domain $S^1$ as four quarter-circles, and the attaching map says that on the first quarter-circle we trace $a_1$ forwards, then on the second quarter-circle we trace $a_2$ forwards, on the third quarter-circle we trace $a_1$ backwards, and on the fourth quarter-circle we trace $a_2$ backwards. This gives $X$ as the torus.
For the labeling scheme $aa$, $X^1$ is a single circle. The domain $S^1$ is two half-circles, and the attaching map traces $a$ on the first half-circle and then $a$ on the second half-circle, forwards both times. This gives $X$ as $\mathbb{R}P^2$.
To see that this results in a surface equivalent to what you'd expect from the labeling scheme, the point is that the wedge-of-circles is equivalent to drawing a (non-filled-in) polygon where each vertex will be identified, and the "filling in the polygon then gluing the edges according to the labeling scheme" is the same thing as "attaching a disc where the boundary circle traces the labeling scheme's rules." (Also, my notation is suggestive of CW complexes and that's exactly what I'm describing here--the CW decomposition of surfaces.)
For a surface $X$ constructed this way, van Kampen's theorem shows that the fundamental group of $X$ is the group whose generators are the circles $a_1, \ldots, a_n$ and relation given by the labeling scheme. This result you are familiar with since it's what your question is about.
So what's going on with $S^2$? Remember that labeling schemes only give surfaces when they satisfy certain properties, and one of the rules for a labeling scheme such that the above consequence of van Kampen's theorem is true is that it's a fully reduced word in the letters $a_1,\ldots,a_n$ (there are more rules about what orders the letters can come up in, how many times each letter shows up, etc., and perhaps these are part of what a 'labeling scheme' is defined as). So $S^2$'s labeling scheme really should be $1$ if we want to use the result, which would then say that the fundamental group is trivial.
Further, $aa^{-1}$ does not in fact produce $S^2$. Otherwise, this would give $S^2$ as a CW complex with a single 0-cell, a single 1-cell, and a single 2-cell. This has Euler characteristic $1$, but the Euler characteristic of $S^2$ is $2$. This is one example of why your labeling scheme needs to be a reduced word.
This raises an important question, which is "can we write $S^2$ as the result of the above process where we start with a wedge of 0 circles?" After all, my description should work if $n=0$ and if it's going to work for all surfaces, and if the labeling scheme for $S^2$ would be something like "$1$"; i.e., start with no circles. The answer is yes: a wedge of zero circles is a single point (because a wedge sum of circles is the same thing as taking the one-point space and then adding copies of $D^1$ according to the constant attaching map $S^0 \to \{ * \}$), so $X^1$ is the one-point space, and then attach $D^2$ to this via the constant map $S^1 \to \{ * \}$; this gives $S^2$.
The other consideration worth mentioning is that if you do start with $X^1 = S^1$ and attach via $aa^{-1}$, we see that this attaching map is nullhomotopic ($a$ can be thought of as a generator for the fundamental group of $X^1$, and $aa^{-1}$ is the group identity in $\pi_1(X^1)$). And it's a fact of adjunction spaces that the homotopy type of the attaching map is all that matters. So, what do you get if you attach a $2$-cell to $S^1$ via a constant map? This is the space $S^1 \vee S^2$.
