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quotient group of a topological group by
its identity component is totally
disconnected∗
joking†
2013-03-22 2:29:45
Assume that G is a topological group and Ge is the identity component. It
is well known, that Ge is a normal subgroup of G, thus we may speak about
quotient group.
Proposition. The quotient group G/Ge is totally disconnected.
Proof. First of all note that connected components of G are of the form gGe .
Indeed, Ge is a connected component of e ∈ G and for any g ∈ G we have a
homeomorphism fg : G → G such that fg (x) = gx. Thus fg (Ge ) = gGe is a
connected component of g ∈ G (please, see this entry for more details).
Now let π : G → G/Ge be the quotient map (which is open and onto) and
A ⊆ G/Ge be an arbitrary, connected subset of G/Ge . Assume that there
are at least two points in A. Consider the subset π −1 (A) ⊆ G (which is the
union of some cosets). Since A has at least two points, then π −1 (A) contains
at least two cosets, which are connected components of G. Thus π −1 (A) is not
connected. Therefore there exist U, V ⊆ π −1 (A) such that U, V are open (in
π −1 (A)), disjoint and U ∪ V = π −1 (A).
Note that if x ∈ U , then the connected component of x (which is equal to
xGe ) is contained in U . Indeed, assume that xGe 6⊆ U . Then there is h ∈ xGe
such that h 6∈ U . Then, since U ∪ V = π −1 (A) we have that h ∈ V . But
then U ∩ xGe and V ∩ xGe are nonempty open and disjoint subsets of xGe such
that (U ∩ xGe ) ∪ (V ∩ xGe ) = xGe . Contradiction, because xGe is connected.
Analogusly, whenever x ∈ V , then xGe ⊆ V .
Therefore both U and V are unions of cosets. Thus π(U ) and π(V ) are
disjoint. Furthermore π(U ) ∪ π(V ) = A and both π(U ), π(V ) are open in A
(because π is an open map). This means that A is not connected. Contradiction.
Thus A has at most one element, which completes the proof. ∗ hQuotientGroupOfATopologicalGroupByItsIdentityComponentIsTotallyDisconnectedi
created: h2013-03-2i by: hjokingi version: h41539i Privacy setting: h1i hTheoremi h22A05i
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Remark. This proposition can
S be easily generalized as follows: assume
that X is a topological space, X = Xi 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 ∈ Xi ). Then, if the quotient
map π : X → X/R is open, then X/R is totally disconnected.
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