Are Quotients by Homeomorphic Subspaces Homeomorphic?by WWGD Tags: homeomorphic, quotients, subspaces 

#1
Nov1912, 12:01 AM

P: 393

Hi, All:
Let X be any topological space, and let A,B be subspaces of X that are homeomorphic to each other. Does it follow that the quotients X/A and X/B are homeomorphic? I know this is true if A,B are both contractible in X , since we then have X/A ~X ~X/B But I'm not sure otherwise. Any ideas? Thanks in Advance. 



#2
Nov1912, 09:54 AM

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so try a case that is not contractible, like two circles on a torus, one contractible and one going around a handle. or just take two circles in a closed disc, one the border of the disc and another one anywhere else, like concentric with that border, but with half the radius.
I don't even believe the contractible case. take a closed disc and attach an interval with one end point of the interval attached at 12 o'clock on the disc and the other at 6 o'clock. then contract the whole disc to get a circle, something one dimensional, or contract a homeomorphic smaller disc to get something homeomorphic to the original space, hence two dimensional. even simpler, just contract a disc to a point, i.e. X = A = the disc, or contract a smaller homeomorphic concentric disc and get a disc. why do you believe the contractible case? are you reading this somewhere? if so, do you know the difference between homeomorphism and homotopy? and are you missing some hypotheses maybe? just asking. 



#3
Nov1912, 08:37 PM

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If there is a homeomorphism that carries A homeomorphically onto B it seems plausible that it would work.




#4
Nov2012, 05:30 PM

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Are Quotients by Homeomorphic Subspaces Homeomorphic?
do you think that is what he meant? since that case is not just plausibly but patently true, the question would hardly arise.




#5
Nov2112, 07:12 PM

P: 393

You're right, it may be that if A homeo. to B, then X/A homotopicallyequivalent to X/B , tho maybe not, given the examples of the torus that you described, i.e., two disks, one homologicallytrivial (i.e., contractible in the torus) , and the other one not trivial.I will continue looking. I don't know if lavinia's condition, which comes down to isotopy is strongenough. I wonder too, if the result would be true for homologicallytrivial spaces. 



#6
Nov2112, 09:52 PM

P: 1,044

What you want for that to be true is a homeomorphism that restricts to a homeomorphism on the subspaces. An arbitrary homeomorphism from A to B might not extend to a homeomorphism of the bigger spaces.




#7
Nov2112, 11:58 PM

P: 393

Right, that is the condition of isotopy we were talking about. I think if the space is
homologically trivial , or unknotted in a given dimension ( I don't know if this is the correct terminology; I mean if any two embeddings of a subspace into the space are isotopic ), the result may be true. There is a version of knottedness from the perspective of extending homeomorphisms, e.g., S^1 is unknotted in R^2 , since any homeomorphism h: (S^1 <R^2 )>(S^1 >R^2) extends into a homeomorphism of R^2 to itself, but this does not generalize into S^2 in R^3 , by, e.g., the Alexander Knotted Sphere. 



#8
Nov2212, 10:45 AM

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i recommend you try to prove as an exercise that lavinia's condition is sufficient. that should be a simple test of whether you understand the concepts. until you can do that, I suspect you are wasting your time memorizing statements and vocabulary.




#9
Nov2212, 11:17 AM

P: 393

I seriously don't see what makes you think I'm just "memorizing terms and vocabulary", specially when only one post ( one of yours ) offered more than
a guess. Why aren't you claiming that the others who just speculated were also just memorizing? I don't expect everyone to know every answer; I sure don't, but how are comments speculating on the answer somehow "more grown up" than mine? Again, nothing wrong in speculating, but somehow I'm just throwing big terms around? An isotopy between subspaces A,B of X is an automorphism of X that takes A to B. Knotedness is equivalent to the problem of extending maps from subspaces into the whole space; is it not? What do I not understand here? Please convince me how the terms I'm using are outofcontext or are being misused. 



#10
Nov2212, 11:53 AM

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you seem to know the definitions but not how to make the simplest use of any of them. that is indicative of memorizing withoutn working out examples and exercises. i used to memorize like that a lot myself, and it got me in trouble. that's how i recognize it and advise against relying on it. try my exercises and outline in your other post on the mapping torus. i hope they help.




#11
Dec112, 12:44 AM

P: 393

Just curious as to a more pointsettheoretical approach to showing if there is
a homeo. from X to X taking A to B, then X/A ~ X/B (or, equiv. a deformation or homotopy of X where the maps are isomorphisms and not just continuous ): Would you consider then, re the above, a collection X/A_t of quotients ; where t in [0,1], A_0:=A and A_1=B to do the job? I mean, we start with X/A :=X/A_0 and then continue by a path of homeomorphisms A_t in (0,1), until reaching A_1=X/B. Then X/B is a composition of the collection of homeomorphisms indexed by t , and so X/A homeo X/B ? 



#12
Dec612, 07:51 AM

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You got a homeo F:X>X that take A to B.
Under what conditions does a continuous map f:X>X descends to a continuous map X/A > X/B ? Well, certainly we need that f(A) lands in B otherwise the map X/A > X/B is illdefined. But actually that is all we need because continuity is taken care of automatically by definition of the quotient topology. Therefor, from this simple observation, our homeo F descends to a homeo X/A > X/B. Similarly, if F_t is a path of homeo taking A to A_t, then at every stage, F_t descends to a homeo X/A > X/A_t. 


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