0708-1300/Class notes for Thursday, January 10
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.
Recall from last class we are going to prove the following lemma:
Then every path with has a unique lift such that
A covering map is a map such that
0) B is connected and locally connected
1) The map is locally a map where F the "fiber" is a discrete set
More precisely this condition means that every point in B has a neighborhood U such that is for some such F
Note 1: As is assumed in this section of the course, our map p is continuous, our spaces have base points . Furthermore we mean, as always, pathwise connected when we say connected.
Note 2: Recall that connected does NOT imply locally connected. Indeed recall the Cantor Comb from a previous class.
Examples of covering maps
1) is trivially a covering map
2) The map from last class,
3) where the identification glues antipodal points on the sphere. Here . I.e., there are two preimages of each point (the two antipodal points)
4) Likewise, etc...
5) Consider a bunch of identical floors of "abstract" parking garages all on top of each.
Recall SO(3) the group of orientation preserving rigid rotations of .
(if then )
This is because we know that any rotation occurs about an axis. So one can encapsulate the information of a given rotation by a vector where the orientation of the vector denotes the axis and the length of the vector denotes the amount of rotation. However, a rotation by degrees does not matter which direction v or -v we do this in, hence the identification. But the resulting set is precisely .
We can now generalize Lemma 1 with our new concept of covering maps opposed to just the map e (which we now know is a specific example of a covering map)
Lemma 1 (general):
Let be a covering map. Then every path with has a unique lift X such that
Cover B with good sets (where by "good" we mean the inverse images look like products)
Now is covering by these and so [0,1] is covered by
The Lebesgue Lemma then implies that there exists N such that [i/N, (i+1)/N] such that is in one of the
The lemma then follows by an inductive argument which is essentially just bookkeeping. Very loosely it is as follows: In the interval [0,1/N] we know that gamma at 0 is and have to uniquely define . We thus get in this interval recalling that for continuity we have to keep in one "level". We thus proceed inductively through the other subintervals of [0,1] and it remains to be checked that everything does in fact work out as we expect.