# Using the Seifert-van Kampen theorem to calculate the fundamental group of the torus, aided by illustrations from mspaint

One of my favorite theorems is the Seifert-van Kampen theorem. It’s a very handy result in algebraic topology which allows us to calculate the fundamental group of complicated spaces by breaking them down into simpler spaces. The version of the theorem I’ll be using here can be stated as follows:

$\textbf{Theorem (Seifert-van Kampen):}$ Let $X$ be a path connected topological space, and let $U_1, U_2$ be subsets of $X$ such that $U_1, U_2$ are both open and path connected, $U_1 \cap U_2$ is path connected, and $U_1 \cup U_2 = X$. If for some $x_0 \in U_1 \cap U_2$ we have

$\pi_1(U_1, x_0) \cong \langle a_1, \dots, a_n \mid R_1 \rangle$

$\pi_1(U_2, x_0) \cong \langle b_1, \dots, b_m \mid R_2 \rangle$

$\pi_1(U_1 \cap U_2, x_0) \cong \langle c_1, \dots, c_j \mid R_3 \rangle$

where $R_1, R_2, R_3$ are a set of relations, then we can calculate the fundamental group of $U_1 \cup U_2 = X$ in the following way. If $I_1:U_1\cap U_2\to U_1$ and $I_2:U_1\cap U_2 \to U_2$ are the inclusion maps, then we have

$\pi_1(U_1 \cup U_2, x_0) \cong \langle a_1, \dots a_n, b_1, \dots, b_m \mid R_1, R_2, R_4\rangle$

where $R_4 = \{I_1(c_k) = I_2(c_k) \mid k\in \{1,\dots, j\}\}$

$\textbf{Important corollary:}$ If $U_1 \cap U_2$ is simply connected, then the fundamental group of $X$ is the free product of the fundamental groups of $U_1$ and $U_2$. This follows by noting that $I_i(C_k) = 1$ for all $i \in \{1,2\}, k \in \{1,\dots, j\}$, and hence $R_4$ consists only of the trivial relation.

Side note: from now on, because $X$ is path connected, the choice of base point does not affect the isomorphism class (this is a classic result). As such, I will be simplifying notation by referring to the fundamental group of $X$ as simply $\pi_1(X)$.

We will use this to calculate $\pi_1(T)$, there $T$ is the torus, which is realized as the following identification space:

We will let $U_1$ be a small disk in the center, and we will let $U_2$ be the complement of $U_1$, plus a little bit, so the intersection will be an open annulus. Here is an illustration:

We can see that $U_1$ is simply connected, so $\pi_1(U_1)$ is trivial. Further, because the intersection of $U_1$ and $U_2$ is an annulus, it has fundamental group isomorphic to $\mathbb{Z}\cong\langle a\mid \emptyset \rangle$, where $a$ is a clockwise loop which encircles $U_1$. We next need to find the fundamental group of $U_1$, which is the punctured torus.

First, we take the “hole” at the center of the square and “stretch” it out:

$\to$

Now, as in the statement of the Seifert-van Kampen theorem above, we will let $U_1$ be defined as follows:

Similarly, here is $U_2$

So, here is $U_1 \cap U_2$

We see that $U_1$ and $U_2$ are both annuli, and hence have fundamental group $\mathbb{Z}$. Further, their intersection is in an X shape (as in, shaped like an uppercase X), so is simply connected. Thus, the fundamental group of the punctured torus is the free product of $\mathbb{Z}$ with itself, which is isomorphic to the free group on two generators.

As a side note, we can specify what the generators of this group are. A loop which starts at the bottom left of the square and travels along the left edge is not homotopic to a loop which starts at the bottom left and travels along the bottom edge.

So, back to this:

To summarize so far, we have:

$\pi_1(U_1) \cong \langle \mid e = 1 \rangle$

$\pi_1(U_2)\cong \langle a, b \mid \emptyset \rangle$

$\pi_1(U_1\cap U_2)\cong \langle c \mid \emptyset \rangle$

We know the fundamental groups of all the relevant spaces, so we are almost done. All we have to do is calculate $R_4$, and we can put all the pieces together. We see that $I_1(c) = 1$. What is $I_2(c)$?

Here is what $I_2(c)$ looks like as a loop in $U_2$. We make the looper wider, pressing it towards the boundary of the square as follows:

$\to$$\to$

The loop is clockwise, so when it reaches the edge, we can trace the loop by going up along the left edge, right along the top edge, down along the right edge, and left along the bottom edge. We noted before that if the fundamental group of the punctured torus is $\langle a, b \mid \emptyset \rangle$, then $a$ can be a loop starting at the bottom left and traversing the left edge, and $b$ can be a loop starting at the bottom left and traversing the bottom edge. Because this is a quotient space, we can see that $a$ is also the same as starting at the top right corner, going “up” to reach the bottom right corner, and traversing the right edge, and $b$ is the same as starting at the top right corner, going “right” to reach the top left corner, and traversing the top edge.

So, $I_2(c)$ is a loop traversing the edges of the square in a clockwise manner. In other words, $I_2(c)$ can be achieved by traversing $a$, followed by $b$, followed by $a^{-1}$, followed by $b^{-1}$. So, $I_2(c) = aba^{-1}b^{-1} = [a, b]$.

So, here’s the punchline:

$\pi_1(T) \cong \langle a, b, e \mid e = 1, I_1(c) = I_2(c) \rangle \cong \langle a, b \mid ab = ba \rangle$

In other words, $\pi_1(T) \cong \mathbb{Z}^2$!