Answer: Connected Sum of Two RP²s and Its Fundamental Group

Key Idea / Intuition

Each $\mathbb{RP}^2$ has $\pi_1 \cong \mathbb{Z}/2\mathbb{Z}$, generated by a loop $a$ satisfying $a^2 = 1$. When you take the connected sum, you glue two copies together along a boundary circle. Van Kampen says the fundamental group is the free product of the two copies, amalgamated over the boundary loop. The surprise is that the boundary circle in each $\mathbb{RP}^2$ represents the element $a^2 = 1$ (it's null-homotopic after the loop goes around twice), so the amalgamation is over the trivial subgroup — giving a free product $\mathbb{Z}/2 * \mathbb{Z}/2$. And this group is isomorphic to the infinite dihedral group, which is the fundamental group of the Klein bottle.

---

Formal Proof / Solution

Setup via Van Kampen

Write $M = \mathbb{RP}^2 \# \mathbb{RP}^2$. Decompose $M$ as: $M = U_1 \cup U_2$ where $U_i \simeq \mathbb{RP}^2 \setminus \{\text{open disk}\}$, and $U_1 \cap U_2 \simeq S^1$ (an open annulus, homotopy equivalent to $S^1$).

Key fact: $\mathbb{RP}^2 \setminus \{$open disk$\}$ deformation retracts onto a Möbius band, whose fundamental group is $\mathbb{Z}$, generated by the core circle $a$.

The boundary $\partial(\text{Möbius band}) = S^1$ wraps twice around the core, so the inclusion $S^1 = U_1 \cap U_2 \hookrightarrow U_i$ sends the generator of $\pi_1(S^1)$ to $a_i^2 \in \pi_1(U_i) \cong \mathbb{Z}$.

Applying Van Kampen

$\pi_1(M) = \pi_1(U_1) *_{\pi_1(U_1 \cap U_2)} \pi_1(U_2) = \mathbb{Z} *_{\mathbb{Z}} \mathbb{Z}$

where the amalgamation is: the generator $t \in \pi_1(S^1) \cong \mathbb{Z}$ maps to $a_1^2$ in $U_1$ and $a_2^2$ in $U_2$.

So: $\pi_1(M) = \langle a_1, a_2 \mid a_1^2 = a_2^2 \rangle$

Part (b): Identification as the Klein Bottle

By the classification of compact surfaces, $\mathbb{RP}^2 \# \mathbb{RP}^2$ is the Klein bottle $K$.

The Klein bottle has the polygon word $abab^{-1}$, giving: $\pi_1(K) = \langle a, b \mid abab^{-1} = 1 \rangle = \langle a, b \mid aba^{-1} = b^{-1} \rangle$

This is indeed isomorphic to $\langle a_1, a_2 \mid a_1^2 = a_2^2 \rangle$ via $a = a_1$, $b = a_1 a_2^{-1}$ (one can verify the relation transforms correctly).

Part (c): Is it Abelian?

No. $\pi_1(K) = \langle a, b \mid aba^{-1} = b^{-1}\rangle$ is non-abelian: $ab \neq ba$ because $ab = ba$ would require $b = b^{-1}$, i.e., $b^2 = 1$, which is not forced by the relation.

In fact, note: $\pi_1(\mathbb{RP}^2 \# \mathbb{RP}^2) \cong \mathbb{Z}/2 * \mathbb{Z}/2$

(the infinite dihedral group $D_\infty$), which has presentation $\langle r, s \mid r^2 = s^2 = 1\rangle$. This is non-abelian and infinite, despite each factor being $\mathbb{Z}/2$.

The surprise: Two finite groups ($\mathbb{Z}/2$) glue together to form an infinite non-abelian group. The connected sum "unkills" the relation $a^2=1$ from each copy — the boundary gluing only forces $a_1^2 = a_2^2$, not $a_i^2 = 1$.

Summary Table

| Space | Fundamental Group | Abelian? | Finite? | |---|---|---|---| | $\mathbb{RP}^2$ | $\mathbb{Z}/2$ | Yes | Yes | | $\mathbb{RP}^2 \# \mathbb{RP}^2$ (Klein bottle) | $\mathbb{Z}/2 * \mathbb{Z}/2 \cong D_\infty$ | No | No |