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Homework Help: Prove there is no homeomorphism that makes two functions conjugate

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    let f(x) = x3 and g(x) = x - 2x3. Show there is no homeomorphism h such that h(g(x)) = f(h(x))

    2. Relevant equations

    Def let J and K be intervals. the function f:J->K is a homeomorphism of J onto K if it is one to one, onto, and both f and its inverse are continuous

    Def let f:J->J and g:K->K, then f is conjugate to g if there is a homeomorphism h:J->K such that h(f(x)) = g(h(x)). (Then use the inverse of h as the homeomorphism that makes g conjugate to f, as in the given question)

    Thm. let f conjugate to g by h. then:
    1. h(fn) = gn(h) for n >=1
    2. if x* is a periodic point of f, then h(x*) is a periodic point of g
    3. if f has a dense set of periodic points, so does g

    Another question in the book goes: let f(x) = x3 and g[itex]\mu[/itex](x) = x - [itex]\mu[/itex]x3. Show there is no nontrivial polynomial h such that f is conjugate to g by h.

    3. The attempt at a solution

    I know nothing of topology so this question is difficult for me to start.

    I want to show there is no h(x) such that h(x)^3 = h(x-2x3). this is from the definition/question statement, but doesnt really seem useful in helping me solve the problem.

    My next intuition is that certain properties are preserved by conjugacy, so if I could show that one of them is violated then I would have a solution.
    for example, periodic points are preserved by conjugacy, so if I can find a periodic point x* of g(x) for which h(x*) is not a periodic point of f(x), then i'd be in business. I used maple to determine that g(x) has period 2 points (1 and -1) and no period 3 points. I also know that 1 and -1 are periodic points of f(x). Can I use this in any way? The trouble I have here is that I dont know what h is going to "do", so I cant really use the theorem I quoted to claim that h(x*) wont be a periodic point for f(x).

    Any ideas would be excellent. I found this other thread on proving no homeomorphism exists (https://www.physicsforums.com/showthread.php?t=423411&highlight=no+homeomorphism) but it seems a little different in that theyre showing two spaces arent homeomorphic to one another whereas Im showing that two functions arent conjugate... also much of the language of topology (compactness, connectedness..) is supposedly outside the scope of the course Im taking and not needed to solve this problem at all. so any hints/help?

    thanks alot
  2. jcsd
  3. Feb 28, 2012 #2
    All you have to prove is that h such that h(x)^3=h(x-2x^3) cannot be one-to-one, I think you can easily find a pair of x's that lead to the same h(x) by making use of the fact that x-2x^3 is not one-to-one
  4. Feb 29, 2012 #3
    Thanks! alot simpler than what I was attempting to do.
  5. Feb 29, 2012 #4
    Okay, so I ended up doing this question differently: since its a homeomorphism then it preserves the fixed points, and so h(g(x)) has a certain number of fixed points which implies that f(h(x)) has the same number, but f only has 1 fixed point and h(g(x)) has 3 i think, so thats the contradiction and thus h cannot exist.

    however I still have a question about your method sunjin09: I can find two points that lead to the same h(x), but the thing is they arent different in the domain of h(x). what i mean is, I can find different points that make g(x) = 0 which are x = 0 and +/-(1/sqrt(2)), but putting these into h(x) would give h(0) each time, so yes, im starting with different values (0, 1/sqrt(2), -1/sqrt(2)) but they are all coming down to h(0), so h still maps 0 to one value. how would I show that h isnt one-to-one?

    just as another example, i can find two points that g(x) maps to the same value (cause its not one-to-one), so i can put each of these points into h(g(x)). but say g(x) maps the value to a number a, then we have h(a) each time, so this doesnt really tell us anything about whether h is injective or not. I dont know if youre still subscribed to this thread but if you are I would appreciate some clarification, cause im still confused.

  6. Mar 4, 2012 #5
    Sorry about the late reply. If g(x1)=g(x2), then h(x)^3=h(g(x)) → h(x)=h(g(x))^(1/3) is the same for x1 and x2, because x^(1/3) is one to one, not h(g(x)). This is what I originally meant.
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