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Prove Inverse of a bijective function is also bijective

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that the inverse of a bijective function is also bijective.


    2. Relevant equations

    One to One
    [itex]f(x_{1}) = f(x_{2}) \Leftrightarrow x_{1}=x_{2} [/itex]

    Onto
    [itex] \forall y \in Y \exists x \in X \mid f:X \Rightarrow Y[/itex]
    [itex]y = f(x)[/itex]

    3. The attempt at a solution
    It is to proof that the inverse is a one-to-one correspondence. I think I get what you are saying though about it looking as a definition rather than a proof.

    How about this..

    Let [itex]f:X\rightarrow Y[/itex] be a one to one correspondence, show [itex]f^{-1}:Y\rightarrow X[/itex] is a one to one correspondence.

    [itex] \exists x_{1},x_{2} \in X \mid f(x_{1}) = f(x_{2}) \Leftrightarrow x_{1}=x_{2} [/itex]

    furthermore, [itex]f^{-1}(f(x_{1})) = f^{-1}(f(x_{2})) \Rightarrow f^{-1}(x_{1}) = f^{-1}(x_{1})[/itex] (by definition of function [itex]f[/itex] and one to one)

    kind of stumped from this point on..
    I may want to transfer this post over to the hw section though, I did post to just get a confirmation on my thoughts on bijection but it is now turning into something a bit more specific than that
     
  2. jcsd
  3. Jan 21, 2013 #2

    jbunniii

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    To show that [itex]f^{-1}[/itex] is an injection, we need to show that [itex]f^{-1}(x_1) = f^{-1}(x_2)[/itex] implies [itex]x_1 = x_2[/itex]. What happens if you apply [itex]f[/itex] to both sides of [itex]f^{-1}(x_1) = f^{-1}(x_2)[/itex]?
     
  4. Jan 21, 2013 #3
    My thoughts are that [itex]f^{-1}[/itex] cancels by definition,

    so to begin let [itex]f^{-1}(x_{1}) = f^{-1}(x_{2})[/itex]
    [itex]\Leftrightarrow f(f^{-1}(x_{1})) = f(f^{-1}(x_{2}))[/itex] since [itex]\forall y \in Y[/itex] the function [itex]f^{-1}(y) = x [/itex] for some [itex]x \in X[/itex]
    [itex]\Leftrightarrow x_{1} = x_{2}[/itex]
     
    Last edited: Jan 21, 2013
  5. Jan 21, 2013 #4

    HallsofIvy

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    Yes, that is correct and shows that, because f is injective, so is [itex]f^{-1}[/itex].

    To show that [itex]f^{-1}[/itex] is surjective, you want to prove that, for any x in X, there exist y in Y such that [tex]x= f^{-1}(y)[/tex].

    Given any x in X, let y= f(x). Then use the other direction: [itex]f^{-1}(y)= f^{-1}(f(x))= x[/itex].
     
  6. Jan 21, 2013 #5
    many thanks HallsofIvy
     
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