The Seifert Van Kampen Theorem Example 3

# The Seifert-van Kampen Theorem Example 3

On The Seifert-van Kampen Theorem page we stated the very important Seifert-van Kampen theorem. We will now look at some examples of applying the theorem. More examples can be found on the following pages:

• The Seifert-van Kampen Theorem Example 3

## Example 3

Let $X$ be the Klein bottle. One construction of the Klein bottle is as follows. Take the unit box $[0, 1] \times [0, 1]$. Identify one pair of opposite edges in the same orientation, and the other pair of opposite edges in the opposite orientation to obtain a quotient space (see the illustration below). Use the Seifert-van Kampen theorem to find $\pi_1(X, x)$. Let $U_1$ and $U_2$ be the following subsets:  Then $U_1 \cap U_2$ is: Now observe that $U_1$ is convex and so we have that:

(1)
\begin{align} \quad \pi_1(U_1, x) = \{ 1 \} \end{align}

Furthermore, observe that the fundamental group of $U_1 \cap U_2$ is generated by a single loop $\alpha$ and is infinite cyclic, so:

(2)
\begin{align} \quad \pi_1(U_1 \cap U_2, x) = \mathbb{Z} \end{align}

The only difficult arises from finding the fundamental group of $U_2$. Observe that the square (with the preserved orientations) is a deformation retract of $U_2$. However, the square with the preserved orientations is precisely a bouquet of two circles. We computed the fundamental group of this space in the first example of applying the Seifert-van Kampen theorem and found that:

(3)
\begin{align} \quad \pi_1(U_2, x) = \langle \alpha, \beta : \emptyset \rangle \end{align}

Let $i_1 : U_1 \cap U_2 \to U_1$ and $i_2 : U_1 \cap U_2 \to U_2$ be the inclusion maps as usual so that $i_{1*} : \pi_1(U_1 \cap U_2, x) \to \pi_1(U_1, x)$ and $i_{2*} : \pi_1(U_1 \cap U_2, x) \to \pi_1(U_2, x)$ are the induced homomorphisms. Then by the Seifert-van Kampen theorem we have that:

(4)
\begin{align} \quad \pi_1(X, x) &= \langle \alpha, \beta : \emptyset, \{ i_{1*}(h) = i_{2*}(h) : h \in \pi_1(U_1 \cap U_2, x) \} \rangle \\ &= \langle \alpha, \beta : 1 = \alpha \beta \alpha^{-1} \beta \rangle \end{align}