Showing that a path is a non-trivial element of a fundamental group.

Question

$\mathbb{P}^{2}$ is the quotient of $S^{2}$ given by the relation: x~y if and only if $y=-x$.

Let $p:S^{2}\to \mathbb{P}^{2}$ be the projection and $\alpha$ be the following path $\alpha(t)=(cos(\pi t),sin(\pi t),0)$.

Show that $p \circ \alpha$ represents a non-trivial element of the $\mathbb{P}^{2}$ fundamental group at the point $x_{0}=[(1,0,0)]$.

My approach was showing that the lifting of $p \circ \alpha$, which is $-\alpha(t)$, is not homotopic to the constante path in $x_{0}$. What is the right approach for this?

Answer 1

Choose the point $(1,0,0)$ as $\tilde{x_0} \in S^2$. Then $\alpha$ is trivial in $\pi_1(\mathbb R P^2)$ if and only if $\alpha$ fixes $\tilde{x_0}$ under the monodromy action since $p_*(\pi_1(S^2))$ is the stabilizer of $\tilde{x_0}$.

As you have noted, it does not fix $\tilde{x_0}$.