Connected Component of Identity of Lie Group is Path Connected

Question

How can I show the claim of the title?

I already know it's an open set. And I know an open connected set in $R^n$ is path connected. But how do I do it in a manifold set up?

Answer 1

Yes. Fix a point $p$ in the connected component $G_e$ of the identity element (you can take $p=e$, of course). Now, consider the set$$C=\{g\in G_e\mid\text{there is a path in $G_e$ joining $p$ to }g\}.$$It is not hard to prove that both $C$ and $G_e\setminus C$ are open subsets of $G_e$. So, since $C$ is not empty ($p\in C$) and $G_e$ is connected, $C=G_e$. Since you can joint $p$ to any other element of $G_e$ through a path, $G_e$ is path-connected.