From: wdh@faron.mitre.org (Dale Hall)
Newsgroups: sci.math
Subject: Re: COMPOSITION OF COVERING MAPS
Date: 31 Mar 93 16:22:19 GMT

In article <3599@eagle.ukc.ac.uk> mrw@ukc.ac.uk (M.R.Watkins) writes:
>Suppose A,B and C are manifolds  
>  f:A -> B and g:B -> C are covering maps...
>   Under what circumstances will g.f:A -> C be a covering map??
>
Um, how about always?

Let p: [0,1] = I --> C, with a lift p~:0 |--> A.  Since g: B --> C is
a covering, the lift f.p~ : 0 |--> B of p(0) extends to a unique
lifting of the path p, p': I --> B (i.e., g.p' = p).  Now, p~(0) is a
lifting of p'(0) (since p' is already an extension of the lifting
f.p~(0) of p(0) to B), and since f: A --> B is a covering, there is a
unique lift of the path p' to p~: I --> A, extending p~(0).  This path
p~ lifts p (trivial), takes 0 to p~(0) as desired, and is unique in
these respects, being the unique lift of a unique lift.  

You didn't need A,B,C to be manifolds.
							Dale.

==============================================================================

Newsgroups: sci.math
From: wdh@faron.mitre.org (Dale Hall)
Subject: Re: COMPOSITION OF COVERING MAPS
Date: Wed, 7 Apr 1993 20:15:38 GMT

In article <1993Apr6.125251.26326@nas.nasa.gov> hook@win25.nas.nasa.gov (Edward C. Hook) writes:
>In article <1993Mar31.162219.17289@linus.mitre.org> wdh@faron.mitre.org (Dale Hall) writes:
>>In article <3599@eagle.ukc.ac.uk> mrw@ukc.ac.uk (M.R.Watkins) writes:
>>>Suppose A,B and C are manifolds  
>>>  f:A -> B and g:B -> C are covering maps...
>>>   Under what circumstances will g.f:A -> C be a covering map??
>>>
>>Um, how about always?
>>
>... This doesn't _quite_ prove the assertion, since a covering map is
>    somewhat more than a fibration with unique path lifting -- what this
>    proves is that the composite gf of covering maps is a fibration with
>    unique path lifting. Because manifolds are locally path-connected and
>    semilocally 1-connected, the result follows from:
>
>    Theorem: Every fibration with unique path lifting whose base space is
>    locally path connected and semilocally 1-connected and whose total space
>    is path connected is a covering projection.
>
>    ( This is Theorem 2.4.10 in Spanier's _Algebraic_Topology_ )
>

I agree, of course.  My (somewhat faulty) memory had identified the
path lifting property with the notion of covering spaces, due to the
above theorem from Spanier and the fact that most spaces I have had
interest in have been CW complexes.  Plus, my dog ate my homework.

To atone for the sloppiness, I dug up an example from Spanier that is
a counterexample to the general statement "Um, how about always?" that
I had made earlier.  I owe thanks to Dan Asimov for prodding me into
this exercise.

Define X, and X_n (for each positive integer n) to be the cartesian product of
a countable collection of 1-spheres.  Let X~_n = R^n x X, and let p_n
map X~_n --> X_n by 

p_n:(t_1,...,t_n,x_1,x_2,...,x_k,...)|-->(y_1,...,y_k,x_1,...x_k,...) 

where y_j = exp(2 \pi i * t_j) (i.e., R^n maps to the n-torus T^n in
the standard way).

It's clear that all the maps p_n are covering maps.  In addition, the
mapping of the disjoint union

	\Union p_n : \Union X~_n --> \Union X_n

is a covering map, since it is so on each component.  The base of this
covering projection, \Union X_n, is the product of X with the set of
positive integers, so there is a covering map 

	\Union X_n --> X.

Howwever, the composition

	\Union X~_n --> \Union X_n --> X

is NOT a covering map, since the topology of the infinite product has
as its basic open sets 

	U_1 x U_2 x ... x U_k x S^1 x S^1 x ... x S^1 x ...

where the U_j are open in S^1; that is, only finitely many factors are
allowed to be proper subsets of S^1.  Taking finite intersections of
such things, we still can only have a finite number of "small"
directions (i.e., axes along which the open set does not encompass
the entire range of the S^1 factor) for an open set.  However, the nth 
pre-image under the above composition  \Union X~_n --> X has n
directions along which neighborhoods can be small (i.e., the R^n
directions: the map R^n --> S^1 x ... x S^1 couldn't be a covering if
S^1 had to be viewed with the INdiscrete topology).  Taking the union
over all n, therefore, we find that no open set in X is "evenly covered".

The point is that, downstairs in X, the basic open sets are "slabs" in
most directions, whereas upstairs in \Union X~_n, the nth component
has to be viewed as having n directions along which small
neighborhoods are allowed.  Taking the X~n onely finitely many at a
time causes no difficulties, but once they're all considered
simultaneously, then given any open U \in X,  the mapping finds a
level beyond which the small open sets upstairs don't map
homeomorphically to the open set U.

Another way of looking at the problem is that the space X has the nth
axis (1 x 1 x ... x S^1 x 1 x 1 x ...) shrinking asymptotically with
n, while the space X~n maintains n axes that are infinite in extent
(R^n).  These two features are incompatible if all values of n must be
taken simultaneously.  It also (sort of) explains where the local
connectivity requirements seem to fail:  if infinitely many S^1
factors are smaller than any given \epsilon, then no point has a
neighborhood for which any loops can be shrunk to a point in the
larger space.

						Dale