# Is my calculation of the Fundamental group of the real projective plane correct?

Problem: find the fundamental group of the real projective plane, $${\mathbb{R}P_2}$$, using the Van Kampen theorem.

So, this is our $${CW}$$-complex model for $${\mathbb{R}P_2}$$:

Firstly, we need a nice open cover of $${\mathbb{R}P_2}$$ with path connected intersection. We use $${U=\mathbb{R}P_2\setminus D^2}$$ and $${V=D^2}$$, where $${D^2}$$ is a disk in the center of the above model. We see $${U\cup V = \mathbb{R}P_2}$$ and $${U\cap V = S^1}$$, which is path connected.

Note that for $${U = \mathbb{R}P_2\setminus D^2}$$, we can deformation retract onto the boundary and end up with something homotopically equivalent to the circle, $${S^1}$$:

a generator for the fundamental group of $${\mathbb{R}P_2\setminus D^2}$$ is thus $${A^{-1}B^{-1} := c}$$ (we label this generator as "$$c$$"). Clearly, $${\pi_1(V) = 0}$$. And $${\pi_1(U\cap V) = \pi_1(S^1) = \mathbb{Z}}$$. And so $$\pi_1(\mathbb{R}P_2) = \frac{0 * \mathbb{Z}}{N}$$ what is $${N}$$? Well - it's the normal subgroup generated by $${f(t)g^{-1}(t)}$$, where $${f : \pi_1(U\cap V) \to \pi_1(U)}$$ is the inclusion of $${U\cap V}$$ into $$U$$, $${g : \pi_1(U\cap V) \to \pi_1(V)}$$ is the inclusion into $$V$$ and $${t \in \pi_1(U\cap V)}$$. Since $${\pi_1(V)}$$ is trivial, $${g(t)=1\ \forall\ t \in U\cap V}$$. So we just need to worry about $${f(t)}$$. Take the identity loop that goes anti-clockwise once around $${S^1}$$, i.e:

So we can see under $${f}$$, the $$1$$-loop in $${S^1}$$ is sent to the equivalence class of the loop $${A^{-1}B^{-1}A^{-1}B^{-1} = c^2}$$. So we have now overall $$\pi_1(\mathbb{R}P_2) = \frac{\mathbb{Z}\langle c \rangle}{\{c^2 = 1\}} = \mathbb{Z}_2$$ this is the correct solution, but was my reasoning throughout the solution correct?

• Looks good to me... Apr 23, 2021 at 21:02
• @IgorRivin cool, thank you for taking the time to read it! Apr 23, 2021 at 21:07