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.
Best Answer
You can use the exact sequence of homotopy groups you mention (without knowing the maps) to get the result once you know $\pi_1(SO(3))$. For that you can use the fact that SU(2) double covers SO(3) and SU(2) is simply connected (being diffeomorphic to the 3 sphere).
EDITED to address Meneldur's comment:
Sorry I glossed over the part where you mentioned you wanted an explicit isomorphism. The fibration $$ SO(n) \to SO(n+1) \to S^n $$ gives $$ \pi_2(S^n) \to \pi_1(SO(n)) \to \pi_1(SO(n+1)) \to \pi_1(S^n) $$ is exact. Now for $n \ge 3$, $\pi_2(S^n)$ and $\pi_1(S^n)$ are both trivial. Therefore $\pi_1(SO(n)) \simeq \pi_1(SO(n+1))$ and this isomorphism is via the inclusion $SO(n) \to SO(n+1)$. So once you have the nontrivial loop generating $\pi_1(SO(3))$ you can just include that in higher dimensions to get the generator for all the other ones.
To get such a loop for $SO(3)$, you need to find a path in $SU(2)$ connecting 1 to -1 and then project that to $SO(3)$. Thinking of $SU(2)$ as unit quaternions, such a path is $q:t \mapsto \cos(\pi t) + i\sin(\pi t)$. Now thinking of $\mathbb R^3$ as purely imaginary quaternions, $q(t)$ corresponds to the element in $SO(3)$ sending $p \mapsto q(t)p\bar q(t)$. Now you can write out this map in matrix form to see what the path looks like in terms of $SO(3)$ as 3x3 matrices. Simply including this path using the inclusion into $SO(n)$ for higher $n$ will, by the LES, give us the non-trivial element of $\pi_1$ for higher $n$.
Unfortunately, I am not sure how to determine if a given loop is nullhomotopic.