TITLE: Lie Groups: Identity Component
QUESTION [5 upvotes]: Main Problem
Given a Lie group.
The connected component of the identity is a Lie subgroup:

It is a subgroup.
It is open.

How to check this using topological tools?
Extra Problem
The quotient by the above is the group of connected components:
$$G_e\lhd G:\quad (gG_e)(hG_e)=(ghG_e)=G_{gh}$$
How to check this using topological tools?

REPLY [6 votes]: It is open because every element of a Lie group (or indeed any manifold) has an open neighbourhood which is homeomorphic to $\Bbb{R}^n$ and is in particular connected. 
To see it is a subgroup, denote by $G^\circ$ the connected component of the identity of $G$. The map $$\mu:G^\circ\times G^\circ\to G$$ defined by $$(g,h)\mapsto g\cdot h$$ is continuous, so its image is connected (because $G^\circ\times G^\circ$ is connected). It also contains the identity (because $\mu(1,1)=1$), so it must lie inside the connected component of the identity, i.e. $g\cdot h\in G^\circ$ for every $g,h\in G^\circ$. Similarly (only using the map $\iota:G^\circ\to G$ defined by $g\mapsto g^{-1}$ instead of $\mu$) we can see that $g^{-1}\in G^\circ$ for every $g\in G^\circ$. Thus, $G^\circ$ is a subgroup. (You can also see this way it is a normal subgroup).