TITLE: Fundamental group of $SO(3)$
QUESTION [9 upvotes]: How can I show that the universal cover of $SO(n)$, for $n\ge 3$, is a double cover? And how does that reflect the fact that the fundamental group of $SO(n)$ has two elements? What is the relation between the fundamental group of a topological space and its universal cover? How come that $SU(2)$ is simply connected but $SO(3)$ is not? Thank you!

REPLY [3 votes]: There are advanced techniques that I am not familiar with (e.g. using fiber bundles, differential geometry) to show that $\pi_1(SO(n)) = \pi_1(SO(3))$ for all $n > 3$. Now $SO(3)$ is homeomorphic to $\Bbb{R}P^3$ which by the Van - Kampen Theorem has $\pi_1$ isomorphic to that for $\Bbb{R}P^2$. This of course is known to have fundamental group isomorphic to $\Bbb{Z}/2\Bbb{Z}$.