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 ${\mathbb{R}P_2}$: enter image description here

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, ${S^1}$:

enter image description here

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:

enter image description here

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?

  • 1
    $\begingroup$ Looks good to me... $\endgroup$
    – Igor Rivin
    Apr 23, 2021 at 21:02
  • $\begingroup$ @IgorRivin cool, thank you for taking the time to read it! $\endgroup$ Apr 23, 2021 at 21:07


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