Let $p$ be a point in $\mathbb{P}^n(\mathbb{R})$ and $\Sigma$ the set containing all the projective lines passing through $p$. Given $s\in \Sigma$ we can define a continuous closed path (let's say $\Omega(\mathbb{P}^n(\mathbb{R}),p,p)$ are all continuous paths starting and ending in $p$). The exercize asks is any two lines $s_1,s_2\in \Omega(\mathbb{P}^n(\mathbb{R}),p,p)$ are homotopic which should be true since we can take a projectivity that maps $s_1$ into $s_2$.
But then it asks to use that fact to prove that $s\in \Omega(\mathbb{P}^n(\mathbb{R}),p,p)$ generates the fundamental group of $\mathbb{P}^n(\mathbb{R})$.
This seems strange to me since the fundamental group of $\mathbb{P}^n(\mathbb{R})$ is $\mathbb{Z}/2$ and $s$ i homeomorphic to a circle.
Can you help me? Thank you!
If you already know that the fundamental group is $\mathbb Z_2$ then there is little to prove except that the homotopy class of $s$ is nontrivial, i.e., that a projective line cannot be continuously contracted to a point.
On the other hand the whole point of the exercise seems to be to prove that all closed curves are homotopic to either a point or a projective line, so it seems to me that the assumption $\pi_1=\mathbb Z_2$ is not supposed to be a given.
Finally, the fact that $s$ is homeomorphic to a circle should not be too surprising - the whole definition of the fundamental group of any space depends on homotopy-equivalence classes of closed paths, i.e., homeomorphic images of the circle. Recall that the group operation of $\pi_1$ is defined by glueing together two closed paths, so "generated by" should be interpreted in these terms.