Fundamental group of torus by van Kampens theorem

Question

So I am currently going through some lecture notes where the fundamental group of a torus is calculated by van Kampen's theorem:

The torus is decomposed into its characteristic fundamental polygon and a circle $o$ inside. Clearly, this circle has $\pi_1(o)=0$ and the intersection between the polygon and the circle is the circle.

So by van Kampen's theorem: The fundamental group of my torus is given by $\pi_1(T^2)= \frac{\pi_1(char.poly)}{N(Im \ (i))}$, where $i: \pi_1(o\cap \ char.poly)=0 \rightarrow \pi_1(char.poly)$ is the homomorphism corresponding to the characteristic embedding and $N(Im(i))$ is the normal subgroup induced by the image of this embedding(as a subgroup of $\pi_1(char.poly)$.

Now, there are two things I don't understand: It is claimed that $\pi_1(char.poly)= \pi_1(S^1 \vee S^1)$( I don't see the relationship between this fundamental polygon and 'an eight') and I don't know how to calculate this normal subgroup there. Is there anybody able to help me a little bit?

Answer 1

Perhaps we're thinking of different notions of "fundamental polygons", but I believe the torus $T^2$ is the fundamental polygon $P$ (obtained as the quotient of a square). The two spaces are certainly homeomorphic, if not the same by definition. Thus $\pi_1(T^2)\cong\pi_1(P)$. We'll decompose and apply van Kampen's theorem to $P$. For convenience, we'll call the horizontal edges $A$ and the vertical edges $B$.

We'll decompose $P$ in almost the same way you suggested: fix a point $x_0$ in the middle of $P$, and let $U$ be $P \setminus \{x_0\}$ and $V$ be a small open disk around $x_0$. Then $\pi_1(T^2) \cong \pi_1(P)\cong\big( \pi_1(U) * \pi_1(V)\big)/N$, where $N$ is the subgroup generated by those "words" in $\pi_1(U) * \pi_1(V)$ that represent loops that are actually nullhomotopic (that is, can be shrunk down to points). In particular, think about the "boundary" word $A^{-1}B^{-1}AB$. It is nontrivial in $U$, but we know that it can actually be shrunk down to a point when it lives in $P$.

The image below depicts a deformation of $U$ to the figure eight $S^1 \vee S^1$. What does this imply for $\pi_1(U)$? I'm also happy to provide more hints.

Answer 2

Briefly: First, you need to distinguish between a disk (which is contractible) and a circle, which has $\pi_1(S^1)\cong\Bbb Z$. When you remove an open ball from the torus (the square with identifications on its edges), what's left deformation retracts to the edges of the square. This is $S^1\vee S^1$, because of the identifications: All four vertices are the same point, and the top and bottom edges become one circle and the left and right edges become the other.

The intersection of the disk and the complement of a smaller disk deformation retracts to a circle, and when we include this circle in the larger piece, it wraps (in $\pi_1$) once around the boundary of the square, giving you the word $aba^{-1}b^{-1}$, which generates the commutator subgroup of $\pi_1(S^1\vee S^1)\cong \langle a,b\rangle$.