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Are Quotients by Homeomorphic Subspaces Homeomorphic?

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WWGD
#1
Nov19-12, 12:01 AM
P: 599
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.
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mathwonk
#2
Nov19-12, 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.
lavinia
#3
Nov19-12, 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.

mathwonk
#4
Nov20-12, 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.
WWGD
#5
Nov21-12, 07:12 PM
P: 599
Quote Quote by mathwonk View Post
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.
I remember reading the result from Hatcher's AT, but I could not find it just now.
You're right, it may be that if A homeo. to B, then X/A homotopically-equivalent to X/B ,
tho maybe not, given the examples of the torus that you described, i.e., two disks, one
homologically-trivial (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 strong-enough.
I wonder too, if the result would be true for homologically-trivial spaces.
homeomorphic
#6
Nov21-12, 09:52 PM
P: 1,275
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.
WWGD
#7
Nov21-12, 11:58 PM
P: 599
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.
mathwonk
#8
Nov22-12, 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.
WWGD
#9
Nov22-12, 11:17 AM
P: 599
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 out-of-context or are being misused.
mathwonk
#10
Nov22-12, 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.
WWGD
#11
Dec1-12, 12:44 AM
P: 599
Just curious as to a more pointset-theoretical 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 ?
quasar987
#12
Dec6-12, 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 ill-defined. 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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