I know that $\pi_1(\mathbb{R}P^2)\cong\mathbb{Z}_2$, but in the square model, I get that $\pi_1(\mathbb{R}P^2)=\langle a,b\colon abab\rangle$. These groups must be isomorphic, but I can't find the isomorphism. What is the trick?
$\mathbb{R}P^2$ is the following model
We want to use Seifert van Kampen, so let $U$ be the complement of a point and $V$ an open ball around that point. Then $U$ deformation retracts onto the boundary which is homotopy equivalent to $S^1\vee S^1$. $V$ is simply connected and $U\cap V$ deformation retracts on $S^1$. Then,
$\pi_1(\mathbb{R}P^2)\cong\mathbb{Z}*\mathbb{Z}*_\mathbb{Z} 1$. Let $i:U\cap V\to U$ be the inclusion, Then $\pi_1(\mathbb{R}P^2)\cong\langle a,b\colon i_*(1)\rangle=\langle a,b\colon abab\rangle$
Best Answer
First of all $U$ deformation retracts to the boundary which is homeomorphic to $\mathbb R\mathbb P^1\cong S^1$ and not $S^1\lor S^1$. So by Van Kampen u get $\pi_1(\mathbb R \mathbb P^1)=\mathbb Z/<i_*(\omega)>$ where $\omega$ is the generator of $\pi_1(U\cap V)\cong\pi_1(S^1)\cong\mathbb Z$. But the generator deformation retracts to a path which winds twice around the boundary circle. Thus u get $\pi_1(\mathbb R \mathbb P^1)=\mathbb Z/<i_*(\omega)>=<a|a^2>\cong\mathbb Z_2$