Intuition on fundamental group of $\mathbb R P^2.$

Question

Show that $\pi_1 (\mathbb RP^2) \neq 0$ by demonstrating a loop in $\mathbb R P^2$ which does not lift to a loop in $S^2.$

Consider the two sheeted covering $p : S^2 \longrightarrow \mathbb R P^2,$ which is the quotient map identifying antipodal points of $S^2.$ If I assume that $S^2$ is simply connected then we can argue that for any $z_0 \in \mathbb RP^2,$ the fundamental group $\pi_1 (\mathbb R P^2, z_0)$ acts on the fiber $p^{-1} (z_0)$ of $z_0$ from the right freely and transitively and hence by the virtue of orbit-stabilizer theorem we can conclude that $\left \lvert \pi_1 (\mathbb R P^2, z_0) \right \rvert = \left \lvert p^{-1} (z_0) \right \rvert = 2$ and as $\mathbb Z/2 \mathbb Z$ is the only group of order $2$ upto isomorphism it follows that $\pi_1 (\mathbb R P^2, z_0) \cong \mathbb Z/ 2 \mathbb Z$ and as $\mathbb R P^2$ is path-connected it's fundamental group is independent of the basepoint and hence we can write $\pi_1 (\mathbb R P^2) \cong \mathbb Z/2 \mathbb Z,$ by simply ignoring the basepoint.

But without using the fact that $S^2$ is simply-connected how do I prove that $\mathbb R P^2$ has a non-trivial fundamental group just by demonstrating a loop in $\mathbb R P^2$ which does not lift to a loop in $S^2$ via the two sheeted covering map $p$ we have started with? How to construct such a non-trivial loop in $\mathbb R P^2\ $? Any help in this regard would be warmly appreciated.

Thanks for investing your valuable time on my question.

Answer 1

In general if $p\colon E\to B$ is a covering and $b\in B$ is a basepoint, then if two loops $\alpha$ and $\beta$ are homotopic, their lifts starting at $e\in p^{-1}(b)$ must have the same endpoint. This follows from the fact that if $h\colon I\times I\to B$ is a path homotopy between $\alpha$ and $\beta$ it can be uniquely lifted to a homotopy $H\colon I\times I\to E$ between the lifts of $\alpha$ and $\beta$. But $H(\{1\}\times I )\subset p^{-1}(b)$, thus $H(\{1\}\times I)$ must be a point, since the fibers of $p$ are discrete.

In particular if $e_1,e_2\in p^{-1}(b)$ are distinct points and $\gamma$ a path from $e_1$ to $e_2$, then $p\circ \gamma$ cannot be nullhomotopic. In the case at hand one can choose a path from $\tilde{z_0}$ to $-\tilde{z_0}$ and project it down to $\mathbb{RP}^2$.

Answer 2

We have $z_0 = p(x_0)$ for some $x_0 \in S^2$. Then $p^{-1}(z_0) = \{ x_0, -x_0\}$. Take any path $u : I \to S^2$ such that $u(0) = x_0$ and $u(1) = -x_0$. The path $\bar u = p \circ u : I \to \mathbb RP^2$ is a loop based at $z_0$. It has $u$ as its unique lift starting at $x_0$. But $u$ is not closed, i.e. no loop.

This excludes that $\pi_1(\mathbb RP^2,z_0) = 0$.

In fact, if $p : \tilde Z \to Z$ is a covering projection and $\bar u : I \to Z$ is a loop based at some $z_0 \in Z$ such that $[\bar u] = 0 \in \pi_1(Z,z_0)$, then all lifts of $\bar u$ are closed.

To see this, let $u$ be any lift of $\bar u$.

$[\bar u] = 0$ means that $\bar u$ is homotopic rel. $\partial I = \{0,1\}$ to the constant loop based at $z_0$ which we denote by $\bar c$. Let $h$ be a homotopy rel. $\partial I$ from $\bar u$ to $\bar c$. There exists a lift $H$ of $h$ such that $H(s,0) = u(s)$ for all $s \in I$. Since $h$ maps $C = I \times \{1\} \cup \partial I \times I$ to $z_0$, the lift $H$ maps $C$ into the fiber $p^{-1}(z_0)$ over $z_0$. Since $C$ is connected, we see that $H(C)$ is a connected subset of $p^{-1}(z_0)$. But $p^{-1}(z_0)$ is discrete, thus $H(C)$ must be a one-point subset of $p^{-1}(z_0)$. This shows that $u(0) = H(0,0) = H(1,0) = u(1)$.