Answer: Fundamental Group of RP²

Key Idea / Intuition

Think of $\mathbb{RP}^2$ as a disk $D^2$ with antipodal boundary points identified. The boundary circle, after identification, wraps around twice before closing up — so the single boundary loop becomes a loop of order 2. Van Kampen tells us the fundamental group is $\mathbb{Z}$ (from the disk's interior, which is trivial) amalgamated with $\mathbb{Z}$ (from a neighborhood of the boundary), with the relation that going around the boundary once in the original disk equals the generator squared in $\pi_1$. The result is $\mathbb{Z}/2\mathbb{Z}$.

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Formal Proof / Solution

Step 1: The cell structure / disk model.

Represent $\mathbb{RP}^2$ as a CW complex: take a disk $D^2$ and identify antipodal points on $\partial D^2 = S^1$. Label the boundary circle with the identification: as you traverse $\partial D^2$ once (angle $0$ to $2\pi$), the quotient traces the boundary twice (since antipodal points get identified, the image circle is traversed twice). So the attaching map of the 2-cell is $a^2$, where $a$ is the generator of $\pi_1$ of the 1-skeleton (which is $S^1$ after identification).

Step 2: Decompose using van Kampen.

Let $p \in \mathbb{RP}^2$ be an interior point of the disk. Decompose: $U = \mathbb{RP}^2 \setminus \{p\} \qquad V = \text{small open disk around } p.$

• $V \cong \mathbb{R}^2$, so $\pi_1(V) = 1$ (trivial).
• $V$ is contractible, so $U \cap V$ deformation retracts to a small circle around $p$, giving $\pi_1(U \cap V) \cong \mathbb{Z}$, generated by a small loop $\gamma$ around $p$.
• $U = \mathbb{RP}^2 \setminus \{p\}$ deformation retracts onto the boundary $\partial D^2$ after identification. Since antipodal points on $\partial D^2$ are identified, the boundary becomes $\mathbb{RP}^1 \cong S^1$. So $\pi_1(U) \cong \mathbb{Z}$, generated by $a$ (once around the identified boundary circle).

Step 3: Track the inclusion $\pi_1(U \cap V) \to \pi_1(U)$.

The small loop $\gamma$ around the interior point $p$ is homotopic (inside $U$) to the boundary of the disk. But the boundary of the disk, when traversed once, wraps twice around the identified $S^1$. Therefore the inclusion-induced map sends: $i_*: \pi_1(U \cap V) \to \pi_1(U), \qquad \gamma \mapsto a^2.$

Step 4: Apply van Kampen.

Van Kampen (Corollary 70.4 in Munkres, since $V$ is simply connected) gives: $\pi_1(\mathbb{RP}^2) \cong \pi_1(U) / N,$ where $N$ is the normal subgroup generated by the image of $i_*$, i.e., generated by $a^2$.

Since $\pi_1(U) = \mathbb{Z} = \langle a \rangle$, we quotient by $\langle a^2 \rangle$:

$\boxed{\pi_1(\mathbb{RP}^2) \cong \mathbb{Z}/2\mathbb{Z}.}$

Why $\mathbb{Z}/2\mathbb{Z}$? Intuitively, the non-trivial loop in $\mathbb{RP}^2$ is a path from $x$ to its antipode $-x$ on $S^2$ (a "half great circle"), projected down. Going around twice lifts to a full loop on $S^2$, which is contractible. So the loop has order exactly 2.

Bonus: This is consistent with the fact that $S^2 \to \mathbb{RP}^2$ is the universal cover (since $S^2$ is simply connected), and the deck transformation group is $\mathbb{Z}/2\mathbb{Z}$ (the antipodal map).