Proving that $\pi_0(G)$ is a group for a topological group $G$

Question

I have been asked to prove that $\pi_0(G)$ is a group for a topological group $G$ by first proving that $G_0$ is a normal subgroup of $G$ and then that $\pi_0(G) \simeq G/G_0$ as a set. Here I have let $G_0$ denote the path component containing the unit element of $G$. I was able to prove that $G_0$ is a normal subgroup of $G$. However, I am having a difficult time constructing the bijection between $\pi_0(G)$ and $G/G_0$. Any guidance would be much appreciated.

Answer 1

Let $C$ be a path component of $G$ and let $g\in C$. Then, since $e_G\in G_0$, $g\in gG_0$. Since $G_0$ is path-connected, then so is $gG_0$, and therefore $gG_0\subset C$. By the same argument, $g^{-1}C\subset G_0$, which implies that $C=g(g^{-1}C)\subset gG_0$. So, $C=gG_0$, and therefore $G/G_0$ is the set of the path components of $G$. That is, it is equal to $\pi_0(G)$.