Answer: Fundamental Group of the Klein Bottle

Key Idea / Intuition

The Klein bottle is built by gluing opposite sides of a square, with one pair glued with a twist. We can decompose it into two Möbius bands (or use the standard van Kampen setup: an open "middle strip" and a small open neighborhood of the boundary word). The boundary word $aba^{-1}b$ encodes the identification, and van Kampen's theorem reads off the fundamental group as a group presentation directly from this word.

The surprising punchline: unlike the torus (where $\pi_1 = \mathbb{Z}^2$ is abelian), the Klein bottle's fundamental group is non-abelian — $a$ and $b$ satisfy $aba^{-1} = b^{-1}$, a relation that says conjugating $b$ by $a$ inverts it.

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Formal Proof / Solution

Step 1: The identification space.

The Klein bottle $K$ is the quotient of $I^2 = [0,1]^2$ by:

• $(x,0) \sim (x,1)$ (top/bottom identified the same way — like a cylinder),
• $(0,y) \sim (1, 1-y)$ (left/right identified with a flip).

The boundary word for the single polygon is: $aba^{-1}b$ where $a$ is the horizontal edge and $b$ is the vertical edge (note: the standard torus word is $aba^{-1}b^{-1}$; here one $b$ is unflipped, yielding $aba^{-1}b$).

Step 2: Van Kampen setup.

Let $x_0$ be the image of the center of $I^2$.

Choose the open cover $K = U \cup V$ where:

• $U$ = image of the interior of $I^2$ plus a small collar near the boundary (homotopy equivalent to a circle $S^1$, going around the boundary word),
• $V$ = small open disk neighborhood of the central point, contractible.

More precisely (standard approach):

Let $U$ = $K \setminus \{p\}$ for a point $p$ inside the single face, and $V$ = small open disk around $p$.

• $V$ is contractible: $\pi_1(V) = 0$.
• $U$ deformation retracts onto the 1-skeleton of $K$, which is a wedge of two circles $S^1 \vee S^1$ (the two edge loops $a$ and $b$ after identification). So $\pi_1(U) = \mathbb{Z} * \mathbb{Z} = \langle a, b \rangle$ (free group on 2 generators).
• $U \cap V$ is homotopy equivalent to $S^1$, and its generator maps (via the inclusion into $U$) to the boundary word $aba^{-1}b \in \pi_1(U)$.

Step 3: Applying van Kampen.

Van Kampen's theorem says: $\pi_1(K) = \pi_1(U) *_{\pi_1(U \cap V)} \pi_1(V) = \frac{\langle a, b \rangle * 1}{\text{(image of } \pi_1(U\cap V)\text{)}}.$

The single loop in $U \cap V$ (generator of $\pi_1(U \cap V) \cong \mathbb{Z}$) maps to the boundary word $aba^{-1}b$ in $\pi_1(U)$, and to the trivial element in $\pi_1(V)$. So we add the relation $aba^{-1}b = 1$, giving:

$\boxed{\pi_1(K) = \langle\, a,\, b \mid aba^{-1}b = 1\,\rangle = \langle\, a,\, b \mid aba^{-1} = b^{-1}\,\rangle.}$

Step 4: Non-abelianness.

The relation says $a$ acts on $b$ by inversion: conjugation by $a$ sends $b \mapsto b^{-1}$. This is the defining relation of an infinite dihedral group — except here $a$ has infinite order too, so it is a semidirect product:

$\pi_1(K) \cong \mathbb{Z} \rtimes \mathbb{Z},$

where the action of the second $\mathbb{Z}$ (generated by $a$) on the first $\mathbb{Z}$ (generated by $b$) is by negation $n \mapsto -n$.

Contrast with the torus: $\pi_1(T^2) = \langle a, b \mid aba^{-1}b^{-1} \rangle \cong \mathbb{Z}^2$ (abelian). The single twist in the identification — $b$ instead of $b^{-1}$ in the boundary word — breaks commutativity and makes the Klein bottle's group genuinely non-abelian.