I am new to algebra topology. Question I want to know is what's the fundamental group of nth symmetric product of $\mathbb{R}$?
Can I calulate like this?
$$ \pi_1(\mathbb{R}^n/S_n)=\pi_0(S_n) $$ since $\mathbb{R}^n$ is path-connected and simply connected Lie group. Then $\pi_0(S_n)$ is nontrivial when $n>2$ according to its connectness ? Thanks.
I believe the theorem you are trying to use is the following: Let $X$ be a space and let $G$ be a group acting freely on $X$. Then there is an exact sequence $\pi_1(G)\to \pi_1(X)\to \pi_1(X/G)\to \pi_0(G)\to\pi_0(X)$. Indeed, $G\to X\to X/G$ becomes a fibration, and this induces a long exact sequence in homotopy groups.
Thus, if the action of $S_n$ on $\mathbb R^n$ were free, you could use the long exact sequence and say $\pi_1(\mathbb R^n)=1\to \pi_1(\mathbb R^n/S_n)\to \pi_0(S_n)=S_n\to \pi_0(\mathbb R^n)=*$ is exact, and hence $\pi_1(\mathbb R^n/S_n)\simeq S_n$.
However, the action is not free, since for example the stabilizer of $0\in\mathbb R^n$ is the entire $S_n$. Thus, the action does not work, and Dietrich Burde shows what the actual fundamental group is.