# Fundamental group of $\mathbb{R}P^2$ in 2 models

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

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

## 1 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 $$\pi_1(\mathbb R \mathbb P^1)=\mathbb Z/$$ 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 $$\pi_1(\mathbb R \mathbb P^1)=\mathbb Z/=\cong\mathbb Z_2$$