I want to find a way of proving $\pi_1(\mathbb R \mathbb P^2)=\mathbb Z_2$ without invoking the Van Kampen theorem.

My thoughts: I want to think of the real projective plane as $D^2$ with antipodal points on the boundary glued. It suffices to show that there are only two types of loops up to homotopy since $\mathbb Z_2$ is the only group with two elements. But I find it hard to show that there is only one nontrivial loop up to homotopy.

Any approach without using Van Kampen theorem is appreciated!

  • 3
    $\begingroup$ You can use covering space theory instead, by exhibiting the 2-sphere as the universal cover. $\endgroup$ Commented Sep 12, 2018 at 2:35
  • $\begingroup$ @QiaochuYuan then what's next? $\endgroup$
    – No One
    Commented Sep 12, 2018 at 2:40
  • $\begingroup$ Somehow this can show (roughly) that every loop has order either one or two, but it is still unclear that there are only two types... $\endgroup$
    – No One
    Commented Sep 12, 2018 at 2:53
  • 1
    $\begingroup$ If you can exhibit the universal cover then you’re done: the fundamental group is the group of deck transformations. There are also several other ways to proceed. $\endgroup$ Commented Sep 12, 2018 at 3:02
  • $\begingroup$ $\mathbb{R}P^2$ is the 2-skeleton of $\mathbb{R}P^3$ so the inclusion induces an isomorphism on $\pi_1$. Since $\mathbb{R}P^3\cong SO(3)$, all you need do is figure out the homotopy groups of the rotation group. Of course, depending on who you are, this may, or may not, be easier, but there are certainly intuitive and visual descriptions of $\pi_1SO(3)$ which you may find appealing. $\endgroup$
    – Tyrone
    Commented Sep 12, 2018 at 9:27

1 Answer 1


Prove that $\mathbb{RP}^2 = \frac{\mathbb{S}^2}{\mathbb{Z}_2}$ thinking of $1$ acting as the identity and $-1$ as the antipodal map. Next, prove that the projection $p:\mathbb{S}^2 \rightarrow \mathbb{RP}^2$ is a covering map. Now, you don't have to go through too much of covering spaces to know that whenever you have a covering map $p: E \rightarrow X$, the group $\pi_1(X, x_0)$ acts on the fiber $p^{-1}(x_0)$ as follows:
For $e \in p^{-1}(x_0)$ and $[\sigma] \in \pi_1 (X, x_0)$, $e \cdot [\sigma]:= \sigma_e(1)$, where $\sigma_e$ is the only lifting of the loop $\sigma$ to $E$ starting at $e$.
Now more generally, for an action of a discrete group $G$ on a space $Y$ and $y\in Y$, if we define $G_y$ as the subset of $G$ fixing $y$ (it is a subgroup), then you have a bijection $b:\frac{G}{G_y}\rightarrow G\cdot y$ such that $b \circ q = L_y$ (or $=R_y$), where $q:G \rightarrow \frac{G}{G_y}$ is the natural projection, and $L_y (R_y): G \rightarrow G\cdot y$ is left (right) multiplication by $y$ (the action could be defined as right or left). If $G$ was a topological group then $b$ would be continuous and onto, giving $\frac{G}{G_y}$ to quotient topology.
What happens is that the subgroup of $\pi_1 (X, x_0)$ fixing $e$ is exactly $p_*(\pi_1(E,e))$, where $p_*$ is the map induced by $p$ in the homotopy groups (also not hard to prove, uses uniqueness of liftings with fixed initial condition). Since $\mathbb{S}^2$ has $\pi_1=0$ and this is a two-fold covering, you have a bijection from $\pi_1(\mathbb{RP}^2, \pi(x_0))$ to a set of two elements. The only group with two elements is $\mathbb{Z}_2$.

I recommend you Hatcher's book, or Greenberg's, to review some covering spaces. This technique can be generalized for any space that is quotient by the action of a group, particularly it provides one of the shortest and more comprehensible proofs that $\pi_1(\mathbb{S}^1,*)=\mathbb{Z}$.


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