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Transformations between spaces

  1. Nov 10, 2011 #1
    Is it possible, in general, to have a one-to-one transformation from Rn to Rm for n>m?

    I'm thinking in the context of geometry, where you want to map a bounded region from a higher space to a lower space.
     
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  3. Nov 11, 2011 #2

    chiro

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    This is an interesting question.

    Intuitively since this is bounded, you would think that it is possible, but given the nature of real numbers, I'm not so sure it is.

    It is definitely not possible if m = 0. You can not map an interval onto a point: it is impossible to go from dimension > 0 to 0 and back again.

    My guess is if you wanted to prove this, you should use an inductive argument that shows that you can't go from m to m-1 in terms of one-to-one properties, and then if that is correct, you're finished.
     
  4. Nov 12, 2011 #3

    Deveno

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    i think you can do this:

    imagine you have the peano space-filling curve (well, actually the hilbert one works better)

    (http://en.wikipedia.org/wiki/Space-filling_curve)

    now this map may not be injective, but it IS surjective, so we can define a right-inverse from R2→R that is injective. using some form of the Cantor–Bernstein–Schroeder theorem, we can obtain a bijection between the two.

    i don't think you can do this with differentiable curves, though.
     
  5. Nov 12, 2011 #4

    disregardthat

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    Deveno, this right-inverse won't necessarily be continuous though. I assume this question is about continuous functions.
     
  6. Nov 12, 2011 #5

    Deveno

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    well, if we have to have a homeomorphism, we can't do it. to see this, just remove one point from R2, the injection we are seeking will map to two disconnected sets, but R2-{x0,y0} is connected.
     
  7. Nov 12, 2011 #6

    disregardthat

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    An injection isn't necessarily an homeomorphism. Also, such an injective function need not necessarily be surjective. But, as it turns out, injections from R^n to R^n have open images and are homeomorphisms onto their image by the invarience of domain theorem.

    http://en.wikipedia.org/wiki/Invariance_of_domain

    As R^m can be seen as a subspace of R^n, such an injective function f would be homeomorphic to an open subset of R^m. This set would then be open and connected in R^m (since R^n is connected), thus generating an homeomorphism from R^n to R^m. But this is impossible due to different fundamental groups.
     
  8. Nov 12, 2011 #7

    Deveno

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    yes, i know that a continuous injection need not be a homeomorphism (even a continuous bijection need not be, the inverse also has to be continuous). but, as you point out, we can easily translate our proposed injection into a homeomorphism, so the connectedness should run both ways (i seem to recall, dimly, this (the open mapping property of continuous injection in Rn) involving a fixed point theorem, somewhere. i dunno, i've slept since then).

    i don't understand your remark about fundamental groups, because π1(Rn) is trivial.
     
  9. Nov 12, 2011 #8

    disregardthat

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    You're right, faulty train of thought there. I was thinking about fundamental groups as one remove a certain subset from R^m. It works for m = 1 by removing a point, m = 2 by removing a line I guess, but I don't see how to formalize it for general m. It can actually be seen easily by the invariance of domain theorem however. R^m can be seen as a closed subspace of R^n, so a homeomorphism would imply an injective function from R^n to R^n with image R^m which must be open by the theorem.
     
  10. Nov 12, 2011 #9
    There is no homeomorphism between R^n and R^m for n not equal to m.

    However, they obviously have the same cardinality (ignoring the n=0 case) so there will be bijections between then.
     
  11. Nov 12, 2011 #10
    Oh, and to see the above, just remove a point for R^n and R^m. If they were homeomorphic before, they will be after removing a point too (just precompose the homeomorphism with a homeomorphism which takes the removed point of R^n to the preimage of the removed point of R^m and then restrict).

    But each of R^n and R^m with a point removed are homotopy equivalent to S^{n-1} and S^{m-1} respectively, which aren't homeomorphic for n=/=m (just look at homology groups).
     
  12. Nov 12, 2011 #11

    lavinia

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    You can make this argument even stronger. R[itex]^{m}[/itex] is a subspace of R[itex]^{n}[/itex] so if there were a continuous map that was 1-1 (or even only 1-1 on R[itex]^{m}[/itex]) from R[itex]^{n}[/itex] onto R[itex]^{m}[/itex] then S[itex]^{m-1}[/itex] would be a retract of S[itex]^{n-1}[/itex]. But your homology argument shows that this is impossible.
     
  13. Nov 12, 2011 #12

    disregardthat

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    Regarding the terminology one-to-one and 1-1, does this mean bijective or simply injective?
     
  14. Nov 12, 2011 #13
    1-1 = one- to- one = injective
    onto = surjective
     
  15. Nov 12, 2011 #14
    Sorry Lavinia, could you explain this a little more? For a retract S^{n-1} to S^{m-1}, wouldn't we need the points of S^{m-1} contained in S^{n-1} to be invariant under the induced map? How do we know that the subspace R^m is left invariant under such a homeomorphism?
     
  16. Nov 12, 2011 #15

    lavinia

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    1-1 means injective to me but I also said onto - I hope
     
  17. Nov 12, 2011 #16

    lavinia

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    My idea was to remove the origin so that the inclusion of R[itex]^{m}[/itex] into R[itex]^{n}[/itex] followed by the 1-1 map from R[itex]^{m}[/itex] onto R[itex]^{m}[/itex] induces an isomorphism on the top homology of R[itex]^{m}[/itex] minus a point - so I was a little hasty because this is not exactly a retract - sorry. But the idea is the same.

    I guess once could say that this map when restricted to the m-1 sphere is a continuous bijection and therefore since the sphere is compact, is a homeomorphism onto its image. This shows that the map must induce an isomorphism on homology from the m-1 sphere onto its its image which is itself homeomorphic to an m-1 sphere. But this map is factored through n space minus a point and its m-1st homology is zero.

    Do you think this is wrong? I suddenly got worried. But I think its OK.
     
    Last edited: Nov 12, 2011
  18. Nov 13, 2011 #17
    Alright, can anyone show an example how this can be done?
    Can anyone show how to map the unit disk to an interval, or the unit square to an interval on R?

    Or how about a cube or sphere mapped to R^2?
     
  19. Nov 14, 2011 #18

    Deveno

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    i posted a link in an earlier post. the function is extremely non-differentiable, and it's inverse isn't continuous (imagine taking the points of the unit square, and mapping them onto an infinite number of "cantor-like" sets on the unit interval). it's actually the inverse you're looking for. basically, the idea is this: you take the unit interval, and bend it so many times that it fills space (the unit square, or other region of the plane).

    this gives us a surjection. using the axiom of choice, pick a point in the preimage of any point in the plane region, this gives a bijection between a subset of a real interval and a region of the plane. some possible things to look at:

    Hilbert curve
    Peano curve
    Moore curve
    Dragon curve
    Terdragon curve
    Gosper curve
    Lesbegue curve (this one is "almost differentiable", which is kinda scary)

    one thing i did not know until recently, is that these types of things are actually used to convert higher-dimensional information such as maps or IP addresses, into linear sequences.
     
  20. Nov 14, 2011 #19

    lavinia

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    As my previous post showed, there is no continuous bijection from R^m to R^m with m <n. The same type of arguments apply for the unit disk or a square or a cube or whatever.

    There are continous space filling maps as was previously posted but they are not 1-1.

    There are bijection between any n dimensional region and any m dimensional region, m>0, but these can not be continuous.
     
  21. Nov 14, 2011 #20
    Basically, in summary,

    There are no homeomorphisms between different dimensional spaces.
    For n<m there are continuous maps R^n onto R^m but they will always fail to be one to one.
    In general, of course there are non-continuous bijections between them too.

    But the original question was can there exist a transformation R^n into R^m which is one to one for n>m. I'll assume we mean continuous here. We don't seem to have answered this question- space filling is mapping lower dimensional things to cover higher ones.

    The answer, I think, is clearly that you can't have a continuous map from R^n into R^m for n>m which is one to one. Is there an easy way to see this?
     
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