Answer: Fundamental Group of the Torus via Van Kampen

Key Idea / Intuition

The torus is built from a square by identifying opposite edges. If you cut out a small open disk from the middle of the square (before identification), you get a space homotopy equivalent to a figure-eight $S^1 \vee S^1$. The disk itself (slightly enlarged to stay open) is contractible. Their intersection is a circle. Van Kampen then says: the fundamental group of the torus is the pushout of the two pieces over their common intersection — and the single relation coming from the boundary loop forces $ab = ba$, giving $\mathbb{Z}^2$.

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Formal Proof / Solution

Setup. Represent $T^2$ as the quotient $[0,1]^2 / \sim$ where $(x,0)\sim(x,1)$ and $(0,y)\sim(1,y)$.

Decomposition. Let:

• $U$ = the image of $[0,1]^2 \setminus \{(1/2, 1/2)\}$ under the quotient map (torus minus a point), slightly thickened to be open.
• $V$ = the image of a small open disk around $(1/2, 1/2)$, which is homeomorphic to $\mathbb{R}^2$ and hence contractible.

Their intersection $U \cap V$ is homotopy equivalent to a small circle around the removed point, so $U \cap V \simeq S^1$, giving:

$\pi_1(U \cap V) \cong \mathbb{Z}, \quad \text{generated by a small loop } \gamma \text{ around the puncture.}$

Homotopy type of $U$. The torus minus a point deformation retracts onto the 1-skeleton of the CW structure of $T^2$ — the wedge of two circles formed by the identified edges. Concretely, $U \simeq S^1 \vee S^1$, so:

$\pi_1(U) \cong \mathbb{Z} * \mathbb{Z} = \langle a, b \rangle \quad \text{(free group on two generators)}.$

Here $a$ and $b$ are the two edge loops of the square.

The boundary relation. The boundary of the square, read as a loop in $U$ (going around all four edges with identifications), traces out:

$\gamma = a b a^{-1} b^{-1}$

in $\pi_1(U)$. This is because the square's boundary, with the identifications $(x,0)\sim(x,1)$ and $(0,y)\sim(1,y)$, reads: right edge $a$, top edge $b$, left edge $a^{-1}$, bottom edge $b^{-1}$.

Van Kampen's Theorem. Since $V$ is contractible ($\pi_1(V) = 1$), the theorem gives:

$\pi_1(T^2) \cong \pi_1(U) *_{\pi_1(U\cap V)} \pi_1(V) \;\cong\; \frac{\pi_1(U)}{\text{normal closure of } \iota_U(\gamma)}$

where $\iota_U : \pi_1(U \cap V) \to \pi_1(U)$ sends the generator of $\pi_1(U \cap V) \cong \mathbb{Z}$ to $aba^{-1}b^{-1}$.

Conclusion. We quotient the free group $\langle a, b \rangle$ by the single relation $aba^{-1}b^{-1} = 1$, i.e., $ab = ba$:

$\boxed{\pi_1(T^2) \cong \langle a, b \mid ab = ba \rangle \cong \mathbb{Z} \times \mathbb{Z} = \mathbb{Z}^2.}$

Why this is beautiful. The single relation comes from the boundary of the square — the key topological content of the identification. The algebraic outcome (free group modulo one commutator = abelian) perfectly mirrors the geometric picture: the torus is a product of two circles, and loops in each direction commute because you can slide them past each other on the surface.