Question: What is the fundamental group of the special orthogonal group $SO(n)$, $n>2$?
Clarification: The answer usually given is: $\mathbb{Z}_2$. But I would like to see a proof of that and an isomorphism $\pi_1(SO(n),E_n) \to \mathbb{Z}_2$ that is as explicit as possible. I require a neat criterion to check, if a path in $SO(n)$ is null-homotopic or not.
Idea 1: Maybe it is helpful to think of $SO(n)$ as embedded in $SO(n+1)$ via $A \mapsto \begin{pmatrix} A & 0 \\ 0 & 1 \end{pmatrix}$. The kernel of the surjective map $SO(n+1) \to \mathbb{S}^{n} \subset \mathbb{R}^{n+1}$, $B \mapsto B e_{n+1}$, is then exactly $SO(n)$. Therefore $\mathbb{S}^n \cong SO(n+1)/SO(n)$. The fundamental group of $\mathbb{S}^n$ is trivial. If one knew how the fundamental group of quotients $Y = X / A$ looks like, this could be helpful.
Idea 2: The map $SO(n+1) \to \mathbb{S}^n$ defined above can also be though of as a principal $SO(n)$-bundle. There is an exact sequence of homotopy groups for those bundles, which might provide the result. But the descriptions of the maps in this sequence I found were quite vague.