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Determine whether the given map is an isomorphism

  1. Aug 1, 2009 #1
    Hello,
    I just cracked open this abstract algebra book, and saw a problem I have no idea how to solve.

    Instruction:
    Determine whether the given map [tex]\phi[/tex] is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not? (Note: F is the set of all functions f mapping [tex]R[/tex] into [tex]R[/tex] that have derivatives of all orders.


    Problem:
    [tex]<F, +> with <F, +>[/tex] where [tex]\phi(f)(x) = \int_{0}^{x}f(t)dt[/tex]

    Definition:
    An isomorphism is bijective mapping (one to one, and onto), and also homomorphic.

    Solution:
    The answer is no because F does not map F onto F? I am very bad at this, what does the following show,
    [tex]\phi(f)(0) = 0[/tex]

    Does that show it is not onto? Can someone explain to me why this is not an isomorphism (I am terrible at this- all I know is the definition from wikipedia of bijective mapping and homomorphis- but I cannot apply it to these problems)?
     
  2. jcsd
  3. Aug 1, 2009 #2
    Re: Isomorphism

    Can you find a counterexample to show that phi is not one to one? That is, find two distinct functions in F, call them f and g, such that phi(f) = phi(g)

    Have you tried checking if it is a homomorphism?
     
  4. Aug 1, 2009 #3

    Hurkyl

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    Re: Isomorphism

    It shows that every function in the image of phi has the value 0 at 0.
     
  5. Aug 1, 2009 #4
    Re: Isomorphism

    I can't seem to find a counterexample. I mean if I let [tex]f(t) = t^{2}[/tex] and let [tex]g(t)[/tex] equal to anything else, it would seem [tex]f(t) = t^{2} \neq g(t),[/tex] unless t = 0- but that doesn't make the function equal, it just shows for a particular domain value, the functions are equal.
     
  6. Aug 1, 2009 #5
    Re: Isomorphism

    Do you know what it means for [itex]\phi[/itex] to be onto? It means that for ANY infinitely differentiable function g we can find a function f such that [itex]\phi(f) = g[/itex]. You have shown that [itex]\phi(f)[/itex] must be 0 when evaluated at 0. Are all infinitely differentiable functions 0 at 0? If you can find some function which is not, then you have your counterexample.
     
  7. Aug 1, 2009 #6
    Re: Isomorphism

    I know onto means that for every element in the domain, there must exist some corresponding element in the range. I am confused by the directions,
    Does this mean f will have derivatives of all order, or will g have derivatives of all order?

    Why do I have to show this?
     
  8. Aug 1, 2009 #7
    Re: Isomorphism

    f and g are both in F as [itex]\phi[/itex] is a function from F to F, so both f and g must have derivatives of all orders.

    You don't HAVE to show it, but you mentioned [itex]\phi(f)(0) = 0[/itex] for all f yourself, which is exactly the same as saying that [itex]\phi(f)[/tex] is 0 when evaluated at 0 for all f so I assumed you had shown it and suspected it could be used.

    Maybe this was simply a mixup when you typed in the definition, but that is true for all functions and not really the definition of onto. A function [itex]f : A \to B[/itex] is said to be onto if for all elements b of B we can find some element a of A such that f(a) =b. For your problem it means that for all infinitely differentiable functions g (remember this is the group we are considering) we can find an infinitely differentiable function f such that [itex]\phi(f) = g[/itex]. We don't know what g is as it's arbitrary but we have:
    [tex]g(x) = \int_0^x f(t)dt[/tex]
    which suggests trying to substitute x=0.
    [tex]g(0) = \int_0^0 f(t) dt = 0[/tex]
    Thus we know that for all infinitely differentiable functions g we have g(0)=0, but is this really true? If not then we have a contradiction so [itex]\phi[/itex] can't be onto. Can you find an infinitely differentiable function that is not 0 at 0? (A very simple example will do)
     
  9. Aug 1, 2009 #8
    Re: Isomorphism

    How about we let the function [tex]g = 2t + 1.[/tex]
    Therefore,
    [tex]
    g(0) = \int_0^0 f(t) dt = \int_0^0 [2t + 1] dt = t^{2} + t = 0
    [/tex]

    But it's suppose to not equal 0, so we can produce the proposed contradiction. I don't know why but this is not obvious to me, I can't find the contradiction.
     
  10. Aug 2, 2009 #9
    Re: Isomorphism

    You can't see the contradiction in the following?
    - Let g(x) = 2x + 1
    - Assume [itex]\phi[/itex] is onto.
    - Then there exists some function f such that [itex]\phi(f) = g[/itex]
    - We have, as you showed, g(0)=0.
    - We also have by plugging in x=0: g(0) = 2*0+1 = 1
    - Since g is a function g(0) can't be both 0 and 1 so we have a contradiction.
    - Thus [itex]\phi[/itex] cannot be onto.
     
  11. Aug 2, 2009 #10
    Re: Isomorphism

    Onto (as you defined) simply means for all elements b of B we can find some element a of A such that f(a) = b. In this way of thinking, our b is essentially
    [tex]
    \int_0^x [2t + 1] dt, [/tex]
    and not defined as [tex]2x + 1[/tex] (or is it)? So why do we have to substitute zero into both ?
     
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