How to show that $\pi_1(\mathbb{RP}^2) = \mathbb{Z}_2$?

Question

How to show that $\pi_1(\mathbb{RP}^2) = \mathbb{Z}_2$?

In general is well known that $\pi_1(\mathbb{RP}^n) = \mathbb{Z}_2, ~ n \ge 2.$

But how to show this assertion?

I have a few knowledge about covering spaces...

Answer 1

Given a covering map $p: Y \to X$, the induced map $p_*: \pi_1(Y) \to \pi_1(X)$ is injective, and the right cosets of $p_*(\pi_1(Y))$ in $\pi_1(X)$ are in bijection to the fiber $p^{-1}(x_0)$ if $Y$ is path-connected. The bijection is $$ p_*(\pi_1(Y))a \mapsto \tilde a(1), \text{ where $a$ is a loop at $x_0$ and $\tilde a$ is its lift at $\tilde x_0$} $$ (This assumes that the notation $ab$ means "first the path a, then $b$". If you use the reverse notation, you have to use the left cosets instead).

Now $p: S^2 \to RP^2$ is a $2$-sheeted covering map and $S^2$ is simply-connected. This means the image has two cosets in the fundamental group of $RP^2$, which is thus $Z_2$.

Answer 2

For a "sufficiently nice" action of a group $G$ on a simply-connected, based space $X$, we can lift a loop $\gamma:[0, 1] \to G/X$ to a loop $\tilde \gamma$ in $G$. The map $\gamma \to \tilde \gamma(1)$ then induces a well-defined isomorphism $\pi_1(G/X) \to G$. In this particular case, "sufficiently nice" means "free and properly discontinuous," but you should really study covering spaces before computing $\pi_1(\mathbb{RP^2}) = \pi_1(S^2/\{\pm 1\})$; the proof of the statement above involves some technical lifting properties that are only a small step away from covering spaces anyway.