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Determining whether the map is an isomorphism

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Let F be the set of all functions f mapping R into R that have derivatives of all orders. Determine whether p is an isomorphism of the first binary structure with the second.

    1. [tex]<F, +>[/tex] with [tex]<R, +>[/tex] where p(f) = f'(0)

    2. [tex]<F, +>[/tex] with [tex]<F, +>[/tex] where p(f)(x) = [tex]\int^{x}_{0} f(t)dt[/tex]

    3. [tex]<F, +>[/tex] with [tex]<F, +>[/tex] where p(f)(x) = [tex]d/dx \int^{x}_{0} f(t)dt[/tex]

    4. [tex]<F, \cdot>[/tex] with [tex]<F, \cdot>[/tex] where p(f)(x) = [tex]x \cdot f(x)[/tex]

    2. Relevant equations

    3. The attempt at a solution

    Some ideas I have:

    I think #1 is false, since f=x and f=x+1 can have the same derivative at 0. Isomorphism requires that p be bijective.

    #3 is true. Simplifying gives p(f)(x)=f(t) for all f in F and x in R.

    I'm not so sure about #2, and #4... The book says they are both false, but I don't really understand it. I'd appreciate any hints/suggestions.
  2. jcsd
  3. Sep 14, 2010 #2


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    It says isomorphism, not just bijection. So for number 3 you also need to show that it preserves the binary structure, i.e. p(f+g)=p(f)+p(g)

    For number 4 I don't understand the notation. What is x(x) supposed to be?

    For 2, there are two questions: Is p a bijection, and does p preserve addition? Why don't you try a couple examples for [tex]f(x)[/tex] to get a feel for what p is doing
  4. Sep 14, 2010 #3
    I fixed #4. Sorry about that.

    For #3, p(f) = f(t), so p is an identity, right? So p(f+g) = f(t) + g(t) = p(f) + p(g).
  5. Sep 14, 2010 #4


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    Ok, so let's look at number 2. There are two questions to answer:

    Is p(f+g)=p(f)+p(g)?

    And is p(f) a bijection?
  6. Sep 14, 2010 #5
    p preserves addition, but it's definitely not a bijection... I don't think it's onto. i.e. [tex]p(f)(x)=\int^{x}_{0} f(t) dt \neq x^2[/tex] for any f in F even though [tex]x^2[/tex] is in R...

    Does that make sense?
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