From: Jerome DUBOIS Subject: A Question about Covering Date: Tue, 05 Sep 2000 21:43:49 +0200 Newsgroups: sci.math Summary: [missing] Hi, Can You Help me about the following : Let q : X -> Y and r : Y -> Z be covering maps. What about p = r o p : X -> Z ; Is it a also a covering map? First, If Z is connected and r is a k-fold covering; then Ok, p is a covering ; More generaly, if r^{-1}(z) is finite for all z in Z, then p is also a covering map. But now if it is not the case, what about p = r o p ? * If Z is good enough, ie. if Z has a universal covering space ; I think p is a covering map, but I can't prove that (help!) * Now, if it is not the case. Can anyone give me an example where p is NOT a covering space ? Thank You, Jerome Dubois. ============================================================================== From: Jerome DUBOIS Subject: Re: A Question about Covering Date: Thu, 07 Sep 2000 19:20:54 +0200 Newsgroups: sci.math Ed Hook wrote: > In article <39B54CF5.C5ACB88C@libertysurf.fr>, > Jerome DUBOIS writes: [quote of previous message deleted, through the following --djr] > |> * Now, if it is not the case. Can anyone give me an example where p > |> is NOT a covering space ? > > If r: Y --> Z is a covering map, then > each z in Z has neighborhoods that are > evenly-covered by r. And each point in the > r-fiber of z has neighborhoods that are > evenly-covered by q. You should be able > to use those two statements to prove that > every z in Z has neighborhoods that are > evenly-covered by p = r o q ... note that > every subset of an evenly-covered set is > itself evenly-covered. > > -- > Ed Hook | Copula eam, se non posit > Computer Sciences Corporation | acceptera jocularum. > NAS, NASA Ames Research Center | All opinions herein expressed are > Internet: hook@nas.nasa.gov | mine alone I don't understand your argument. Moreover I find a counter-example when the space Z haven't got any universal covering. (But when the space Z had a universal covering p = r o q is a covering ; but my proof works with a none direct argument.) Ok ; take for Z the infinite shrinking circles : Z = { Cercle[(1/n,0),rayon=1/n] : n > 0 } : Z haven't got any universal covering. Now take for Y the real line with upon each integer the space Z ; one has a covering map r : Y --> Z. Finally one can construct a 2-fold covering q : X --> Y s.t. r o q is Not a covering map (something wrong at point O). Let me know if something was wrong, Thak You JD. ============================================================================== From: hook@nas.nasa.gov (Ed Hook) Subject: Re: A Question about Covering Date: 7 Sep 2000 20:04:06 GMT Newsgroups: sci.math In article <39B7CE76.8B89305F@libertysurf.fr>, Jerome DUBOIS writes: [quote of previous message deleted, through the following --djr] |> I don't understand your argument. Good -- it turns out to be what's known (technically :-) as a "brain fart". What I was thinking (put differently, the mistake that I was making): Take any point z in Z - since r is a covering map, every point in r^{-1}(z) has neighborhoods that are mapped homeomorphically by r. And, inside any such neighborhood, there are neighborhoods that are evenly-covered by q. So, for each point y in r^{-1}(z), choose such a neighborhood N(y). Then (here comes the mistake) \intersect { N(y) | y in r^{-1](z) } is, I was thinking, a neighborhood of z that's evenly-covered by r o q. Well, it's certainly evenly-covered ... but there's no guarantee whatsoever that it's a neighborhood -- it is if r has only finitely many sheets, but not in general. Unless (as you note) Z is nice enough to have a universal cover -- in that case, every covering map is a "factor" of the universal covering map and you can get the evenly-covered neighborhoods needed here by pushing them down from the universal cover. (You will, of course, want to verify that statement for yourself, given my track record in this thread ... ) [quote of remainder of previous message deleted --djr] -- Ed Hook | Copula eam, se non posit Computer Sciences Corporation | acceptera jocularum. NAS, NASA Ames Research Center | All opinions herein expressed are Internet: hook@nas.nasa.gov | mine alone