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Show that if f: A→B is surjective and and H is a subset of B, then f([

  1. Aug 20, 2013 #1
    [f]^{}[/2]1. The problem statement, all variables and given/known data


    Show that if f: A→B is surjective and and H is a subset of B, then f(f^(-1)(H)) = H.

    2. Relevant equations



    3. The attempt at a solution

    Let y be an element of f(f^(-1)(H)).
    Since f is surjective, there exists an element x in f^(-1)(H) such that f(x) = y.
    But x in f^(-1)(H) implies that f(x) is in H, by definition of inverse functions.
    Therefore f(f^(-1)(H)) is a subset of H.

    Conversely, for y in H y must also be in f(f^(-1)(H)) since the latter expression is equivalent to H by definition
    of inverse functions.

    Since f(f^(-1)(H)) is a subset of H and vice versa, f(f^(-1)(H)) = H.
     
    Last edited: Aug 20, 2013
  2. jcsd
  3. Aug 20, 2013 #2

    Office_Shredder

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    This part is good

    This part less so. What if f(x) = 0 for all x a real number, and H is the interval [0,1]? What is f(f-1(H)) in this case?

    You have to use the fact that f is surjective.
     
  4. Aug 20, 2013 #3
    O I understand now I think, thank you :3

    f is surjective, so for any y in H we have f(x) = y.
    since f^(-1)(H) is in the domain of f by definition of inverse functions,
    y = f(f^(-1)(y)) for any y in H and thus H is a subset of f(f^(-1)(H)).

    This line of reasoning checks out, right?
     
  5. Aug 20, 2013 #4

    verty

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    Consider using this type of argument: ##A \subseteq B \land A \supseteq B → A = B##.
     
  6. Aug 20, 2013 #5

    vela

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    You need to be a bit more careful. Consider the case of f: ℝ→{0,∞} where f(x)=x2. The mapping ##f^{-1}## isn't a function, which you seem to be assuming. What would ##f^{-1}(4)## be equal to, for instance — 2 or -2?
     
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