Editing Covering group (section)

== Properties ==
Let ''G'' be a covering group of ''H''. The [[kernel (group theory)|kernel]] ''K'' of the covering homomorphism is just the fiber over the identity in ''H'' and is a [[discrete group|discrete]] [[normal subgroup]] of ''G''. The kernel ''K'' is [[closed set|closed]] in ''G'' if and only if ''G'' is [[Hausdorff space|Hausdorff]] (and if and only if ''H'' is Hausdorff). Going in the other direction, if ''G'' is any topological group and ''K'' is a discrete normal subgroup of ''G'' then the quotient map {{nowrap|''p'' : ''G'' → ''G'' / ''K''}} is a covering homomorphism.

If ''G'' is [[connected space|connected]] then ''K'', being a discrete normal subgroup, necessarily lies in the [[center (group theory)|center]] of ''G'' and is therefore [[abelian group|abelian]]. In this case, the center of {{nowrap|1=''H'' = ''G'' / ''K''}} is given by
: <math>\mathrm{Z}(H) \cong \mathrm{Z}(G)/K .</math>

As with all covering spaces, the [[fundamental group]] of ''G'' injects into the fundamental group of ''H''. Since the fundamental group of a topological group is always abelian, every covering group is a normal covering space. In particular, if ''G'' is [[path-connected]] then the [[quotient group]] {{nowrap|''π''<sub>1</sub>(''H'') / ''π''<sub>1</sub>(''G'')}} is isomorphic to ''K''. The group ''K'' [[Group action (mathematics)|acts]] simply transitively on the fibers (which are just left [[coset]]s) by right multiplication. The group ''G'' is then a [[principal bundle|principal ''K''-bundle]] over ''H''.

If ''G'' is a covering group of ''H'' then the groups ''G'' and ''H'' are [[locally isomorphic groups|locally isomorphic]]. Moreover, given any two connected locally isomorphic groups ''H''<sub>1</sub> and ''H''<sub>2</sub>, there exists a topological group ''G'' with discrete normal subgroups ''K''<sub>1</sub> and ''K''<sub>2</sub> such that ''H''<sub>1</sub> is isomorphic to {{nowrap|''G'' / ''K''<sub>1</sub>}} and ''H''<sub>2</sub> is isomorphic to {{nowrap|''G'' / ''K''<sub>2</sub>}}.