Generators of group $\pi_1(n\mathbb{R}P^2)$ Task:
Prove that the group $\pi_1(n\mathbb{R}P^2)$ is generated by elements $a_1,\ldots, a_n,$ obeying unique relation $a^{2}_{1}\cdot\ldots\cdot a^{2}_{n}=1$.
I know how to solve it, but I think that
$$a^{2}_{1}=1,$$
$$\vdots$$
$$a^{2}_{n}=1.$$
Am I right?
Thanks
 A: No, the problem is correct as stated. You can think of $n\mathbb{RP}^2$ as a $2n$-gon with edges identified in pairs according to the scheme $a_1a_1a_2a_2\cdots a_na_n$ as you go around the $2n$-gon.
Something to think about: $\mathbb{RP}^2\#\mathbb{RP}^2$ is homeomorphic to a Klein bottle. Would your formula make sense for the Klein bottle? 
Edit: In more detail, apply Van Kampen's theorem to the following two open sets: Let $U$ be an open neighborhood of the union of edges, and let $V$ be the open $2n$-gon. Then $U$ deformation retracts to the one point union of $n$ circles, so has fundamental group freely generated by $a_1,\ldots,a_n$. $\pi_1(V)=0$ since the disk is contractible. The intersection of $U$ and $V$ deformation retracts to a circle which reads off the boundary, so $\pi_1(U\cap V)$ is cyclic, generated by $a_1^2\cdots a_n^2$. Van Kampen's theorem then implies that $\pi_1(n\mathbb{RP}^2)\cong \pi_1(U)*_{\pi_1(U\cap V)}\pi_1(V)$, which is $F(a_1,\ldots,a_n)*_{\mathbb Z\langle a_1^2\cdots a_n^2\rangle} \{1\}\cong \langle a_1,\ldots,a_n\,|\,a_1^2\cdots a_n^2\rangle$.
A: I'll use $n P$ as a shorthand for $n \mathbb{RP}^2$, the connect sum of $n$ copies of the real projective plane.  Thus $\pi_1(nP)$ is torsion-free when $n > 1$. So, in general, no, we do not have $a_i^2 = 1$.  
