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For help, A simple question about topology

  1. Jun 3, 2007 #1
    Is a continuous 1-1 and onto mapping from the Euclidean 3-space to itself a homeomorphism? i.e. Is its inverse also continous?
     
  2. jcsd
  3. Jun 3, 2007 #2
    Yes, a function is bijective if it is a continuous 1-1 and onto mapping. If a function is bijective it's easy to show it has an inverse. A homeomorphism is basically a bijection between topological spaces. Since euclidean space is a topological space, it is a homeomorphism.
     
  4. Jun 3, 2007 #3

    morphism

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    A function is bijective if it's 1-1 and onto. That's all - it has nothing to do with continuity. A homeomorphism is a continuous bijection whose inverse is continuous as well. A continuous bijection between topological spaces may as well have an inverse that isn't continuous.

    In calculus it's usually proved that a continuous bijection f:R->R is a homeomorphism. I don't think this generalizes to R^3 though. IIRC, there's a theorem that says if f:R^n->R^n is a continuous bijection that has first-order partials and nonzero jacobian, then its inverse is continuous.
     
    Last edited: Jun 3, 2007
  5. Jun 3, 2007 #4
    Yes, that's it. My error is due to looking at the question and retyping some of it too quickly, sorry.
     
  6. Jun 4, 2007 #5
    In general a continuous bijection with a compact domain will have a continuous inverse.
     
  7. Jun 4, 2007 #6
    What theorem is this? I don't recall it from calculus.
     
  8. Jun 4, 2007 #7

    morphism

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    It usually comes before the inverse function theorem. It's probably stated as: a function that is continuous and 1-1 on an interval has a continuous inverse.
     
  9. Jun 4, 2007 #8

    mathwonk

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    for continuous bijective maps from R to R it is easy, since they are monotone.

    for R^2 it isn't coming to me immediately yet. i think you have to prove the map is proper somehow, if it is.
     
  10. Jun 4, 2007 #9

    mathwonk

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    ok. let p map to q (R^2-->R^2). want to show f^(-1) continuous at q.

    take a disc centered at p, and by jordan curve theorem, the circular boundary maps to a closed loop that has q in its interior.

    thus the image of the compact disc around p is a compact "disc" around q, with q in the interior.

    then the continuous inverse of the restriction of the map to the compact disc, is also continuous as a map on all of R^2.

    so yes.

    whew! what about R^3? I guess th same is true, i.e. the image of a ball around p should map to a ball around q, and use the same argument.

    so this seems to use the jordan theorem to argue that the image of a sphere separates R^3 into an inside and an outside. (provided the map was defined and bijective on all of R^3.)

    i hope this is true. look up the generalized jordan curve theorem. [yep, hatcher, p.169.]

    so this question is simple but difficult.
     
    Last edited: Jun 4, 2007
  11. Jun 4, 2007 #10
    Sorry, delete it. Thanks
     
    Last edited: Jun 4, 2007
  12. Jun 4, 2007 #11
    I know it, a contiunous bijection with a compact domain and a Hausdorff range is a homeomorphism. Unfortunately, here R^3 is not compact.
     
  13. Jun 4, 2007 #12
    In calculus it's usually proved that a continuous bijection f:R->R is a homeomorphism. [/QUOTE]

    I think it can be proved by the following steps:
    step 1: show that f is monotone
    setp 2: a monotone continous function has continouse inverse.

    Is the sketch of proof right? and can you provide a detailed proof of this proposition.
     
  14. Jun 4, 2007 #13
    What is a Jordan curve

    Let C be the circle x^2+y^2=1 in R^2, f is continous bijection from R^2 to R^2, is f(C) a Jordan curve? and what is a Jordan curve? what is its definiton?
     
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