quotient group of a topological group by its identity component is totally disconnected
Proof. First of all note that connected components of are of the form . Indeed, is a connected component of and for any we have a homeomorphism such that . Thus is a connected component of (please, see this entry (http://planetmath.org/HomeomorphismsPreserveConnectedComponents) for more details).
Now let be the quotient map (which is open and onto) and be an arbitrary, connected subset of . Assume that there are at least two points in . Consider the subset (which is the union of some cosets). Since has at least two points, then contains at least two cosets, which are connected components of . Thus is not connected. Therefore there exist such that are open (in ), disjoint and .
Note that if , then the connected component of (which is equal to ) is contained in . Indeed, assume that . Then there is such that . Then, since we have that . But then and are nonempty open and disjoint subsets of such that . Contradiction, because is connected. Analogusly, whenever , then .
Therefore both and are unions of cosets. Thus and are disjoint. Furthermore and both , are open in (because is an open map). This means that is not connected. Contradiction. Thus has at most one element, which completes the proof.
Remark. This proposition can be easily generalized as follows: assume that is a topological space, is a decomposition of into connected components and is an equivalence relation associated to this decomposition (i.e. if and only if there exists such that ). Then, if the quotient map is open, then is totally disconnected.
|Title||quotient group of a topological group by its identity component is totally disconnected|
|Date of creation||2013-03-22 18:45:35|
|Last modified on||2013-03-22 18:45:35|
|Last modified by||joking (16130)|