quotient group of a topological group by its identity component is totally disconnected

Assume that $G$ is a topological group and $G_e$ is the identity component. It is well known, that $G_e$ is a normal subgroup of $G$, thus we may speak about quotient group.

Proposition. The quotient group $G/G_e$ is totally disconnected.

Proof. First of all note that connected components of $G$ are of the form $g{G}_e$. Indeed, $G_e$ is a connected component of $e\in G$ and for any $g\in G$ we have a homeomorphism $f_g:G\to G$ such that $f_g(x)=gx$. Thus $f_g(G_e)=g{G}_e$ is a connected component of $g\in G$ (please, see this entry (http://planetmath.org/HomeomorphismsPreserveConnectedComponents) for more details).

Now let $\pi :G\to G/G_e$ be the quotient map (which is open and onto) and $A\subseteq G/G_e$ be an arbitrary, connected subset of $G/G_e$. Assume that there are at least two points in $A$. Consider the subset ${\pi }^{-1}(A)\subseteq G$ (which is the union of some cosets). Since $A$ has at least two points, then ${\pi }^{-1}(A)$ contains at least two cosets, which are connected components of $G$. Thus ${\pi }^{-1}(A)$ is not connected. Therefore there exist $U,V\subseteq {\pi }^{-1}(A)$ such that $U,V$ are open (in ${\pi }^{-1}(A)$), disjoint and $U\cup V={\pi }^{-1}(A)$.

Note that if $x\in U$, then the connected component of $x$ (which is equal to $x{G}_e$) is contained in $U$. Indeed, assume that $x{G}_e⊈U$. Then there is $h\in x{G}_e$ such that $h\notin U$. Then, since $U\cup V={\pi }^{-1}(A)$ we have that $h\in V$. But then $U\cap x{G}_e$ and $V\cap x{G}_e$ are nonempty open and disjoint subsets of $x{G}_e$ such that $(U\cap x{G}_e)\cup (V\cap x{G}_e)=x{G}_e$. Contradiction, because $x{G}_e$ is connected. Analogusly, whenever $x\in V$, then $x{G}_e\subseteq V$.

Therefore both $U$ and $V$ are unions of cosets. Thus $\pi (U)$ and $\pi (V)$ are disjoint. Furthermore $\pi (U)\cup \pi (V)=A$ and both $\pi (U)$, $\pi (V)$ are open in $A$ (because $\pi$ is an open map). This means that $A$ is not connected. Contradiction. Thus $A$ has at most one element, which completes the proof. $\mathrm{\square }$

Remark. This proposition can be easily generalized as follows: assume that $X$ is a topological space, $X=\bigcup X_i$ is a decomposition of $X$ into connected components and $R$ is an equivalence relation associated to this decomposition (i.e. $xRy$ if and only if there exists $i$ such that $x,y\in X_i$). Then, if the quotient map $\pi :X\to X/R$ is open, then $X/R$ is totally disconnected.

Title	quotient group of a topological group by its identity component is totally disconnected
Canonical name	QuotientGroupOfATopologicalGroupByItsIdentityComponentIsTotallyDisconnected
Date of creation	2013-03-22 18:45:35
Last modified on	2013-03-22 18:45:35
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	7
Author	joking (16130)
Entry type	Theorem
Classification	msc 22A05