Covering transformations
[18/100%]If is a covering map, then a homeomorphism with is called a covering transformation. Suppose is a covering map, and . Suppose that is a covering transformation with . Show that .
Suppose is a covering map with . Show that there is a covering transformation with iff .
Show that the set of all covering transformations of forms a group, which we denote , under the operation of composition of homeomorphisms.
Show that if are covering transformations with for some , then .
Consider the homeomorphism , , . Let be the standard covering map .
Show that .
Show that if are two liftings of a loop , then for we have .
Show that if is a covering transformation, then for some (Exercise 4.3.5). Conclude that for .
Find the group of covering transformations of the covering , (where and denotes the equivalence class of ).
Find the group of covering transformations of the covering , (see Exercise 4.1.3).
Consider the covering space of from Exercise 4.2.8. Show that there is no covering transformation sending to .
Prove the following theorem: Suppose is a covering space with and . Then the group of covering transformations is isomorphic to the fundamental group .
Follow the outline below. Define a map as follows. Given , let be a path in connecting and , and let . Prove that is an isomorphism as follows.
Show that if is homotopic rel to , then .
Use to conclude that any two paths connecting and are homotopic rel and so is well defined.
Show that is a homomorphism.
Show that is --.
Show is onto.
Suppose is a covering space with , . Let . Show that if there is a covering transformation with , then , and if we call this common subgroup , there is an element with , where is represented by a loop which lifts to a path from to .
For the converse, suppose that , , and , where is represented by a loop which lifts to a path from to . Show that there is a covering transformation which sends to .
Use Exercise 4.3.11 to give another argument for Exercise 4.3.9.
The normalizer of a subgroup is the subgroup . Prove the following theorem, which generalizes Theorem 4.3.1 (Exercise 4.3.10): there is an isomorphism between the group of covering transformations of and the quotient of .
Show that there is a bijection between
and (notation as in Exercise 4.3.13).
Suppose that is a normal subgroup of . Then show that for every pair of points in the pre-image , there is a covering transformation with .
Show that the converse is true: whenever it is the case that for every pair of points in the pre-image , there is a covering transformation with , then is a normal subgroup of .
A group of homeomorphisms of a space is called properly discontinuous if for every there is an open set containing such that is disjoint from whenever . Show that if is a group of homeomorphisms which is properly discontinuous, then for every there is an open set containing such that for all .
Recall that a covering space is called regular if is a normal subgroup of ; by Exercise 4.3.15, this is equivalent to the group of covering transformations acting transitively on . Recall also that a group of homeomorphisms of a space is properly discontinuous (Exercise 4.3.16) if for every there is an open set containing such that whenever .
Suppose that is path connected and locally path connected and is a group of homeomorphisms of . Let denote the quotient space formed from , where for .
Show that the quotient map is a covering map iff the action of is properly discontinuous.
Show that when the conditions of (a) are satisfied, then the covering map is regular and is its group of covering transformations.
Prove the following theorem (characterization of regular covering spaces): suppose is a regular covering space (Exercise 4.3.17) and is its group of covering transformations. Then , where is the quotient covering space and is a homeomorphism induced by via the quotient construction.