What needed to be checked for a topological subgroup?

Question

I have the following Problem:

let $G$ be a topological group. We denote by $G_0$ the path connected component of the identity and by $G_0'$ the connected component of the identity. I now want to show that $G_0$ is a normal subgroup of $G$

I wanted to proceed as follows:

Claim 1 $G_0$ is a subgroup of $G$. We need to show the following three points:

1. (Contains the identity element) Indeed we see that by definition $id\in G_0$.
2. (stable under composition) Now I wanted to take two path connected components $A,B\in G_0$. Then clearly $id\in A\cap B$ and thus the intersection is not empty. But then from the lecture we know that $A\cup B$ is again path connected and thus in $G_0$
3. (Stable under inverses) here I have no idea because I don't know how an inverse of a connected subset looks like

Since my approach was wrong I add here a correct one, which was discussed below:

Afterwards I would check that it is normal but for the moment I would be happy if you could take a look at my approach and correct me because I'm really unsure about topological groups since we have never disscused them in class, the only appeared in one exercise sheet.

Thanks for your help.

Answer 1

If $a,b\in G_0$ and $\gamma\colon [0,1]\to G$ is a path from $id$ to $a$, then $[0,1]\to G$, $t\mapsto \gamma(t)b$ is a path from $b$ to $ab$. Append this to a path from $id$ to $b$ to find a path from $id$ to $ab$. Also note that $[0,1]\to G$, $t\mapsto a^{-1}\gamma(1-t)$ is a path from $id $ to $a^{-1}$. Conclude that $G_0$ is a subgroup.

Use a similar simple argument for normality (or perhaps exhibit a homomorphism to a suitable group that has precisely $G_0$ as kernel - this could kill subgroup and normality in a single strike).

Answer 2

The path-connected component of a point $x$ in some space $X$ consists of all points $y\in X$ such that there exists a $xy$-path in $X$. All those $y$ define a subset and thus with the subspace topology a subspace.

A topological group is a topological space $G$ together with a chosen identity element $e_G \in G$ and two continuous maps $\cdot:G \times G \rightarrow G$ and $(-)^{-1}:G \cong G$ satisfying the usual group axioms (associativity of multiplication, unitality of $e_G$ and having inverses given by $(-)^{-1}$).

You are supposed to show that the connected component subspace of the element $e_G$ of a topological group $G$ again forms a topological group. $e_G$ is contained in its own connected component, because there is a constant $e_Ge_G$-path. It is not so obvious that given $x,y$ in the component $x\cdot y$ is in the component. There we need to recall that a $xy$-path is given by a continuous map $[0,1] \rightarrow X$ sending $0$ to $x$ and $1$ to $y$. Since both $x$ and $y$ are in the component of $e_G$ there are $xe_G$- and $ye_G$-paths $\alpha,\beta$. Now the multiplication is a continuous map $G\times G \rightarrow G$ and we can use the paths $\alpha,\beta$ to define a continuous map $$\begin{array}{ccccc} [0,1] & \overset{(\alpha,\beta)}{\longrightarrow} & G \times G &\overset{\cdot}{\longrightarrow} & G\\ t & \mapsto & (\alpha(t),\beta(t)) & \mapsto & \alpha(t)\cdot \beta(t)\\ 0 & \mapsto & (x,y) & \mapsto & x\cdot y\\ 1 & \mapsto & (e_G,e_G) & \mapsto & e_G \cdot e_G = e_G \end{array}$$ This defines a $(x\cdot y)e_G$-path proving that $x\cdot y$ is contained in the component of $e_G$.

I leave the proof of $x^{-1}$ being in the component up to you. The idea is precisely the same: use an $xe_G$-path and continuity of $(-)^{-1}$ to define an $x^{-1}e_G$-path.