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

enter image description here

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 $\pi_1(\mathbb{R}P^2)\cong\langle a,b\colon i_*(1)\rangle=\langle a,b\colon abab\rangle$

  • $\begingroup$ That's not the correct presentation; your second group maps onto $(\mathbb{Z/2})^2$ $\endgroup$ – Max Mar 11 at 11:08
  • $\begingroup$ Perhaps write down your "square model" and find the mistake. $\endgroup$ – Tyrone Mar 11 at 11:15
  • $\begingroup$ I have edited my question $\endgroup$ – user408856 Mar 11 at 11:23
  • $\begingroup$ Is the mistake that $U$ doesn't deformation retract on the bouquet of circles? $\endgroup$ – user408856 Mar 11 at 11:32

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 $\pi_1(\mathbb R \mathbb P^1)=\mathbb Z/<i_*(\omega)>=<a|a^2>\cong\mathbb Z_2$


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