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Homework Help: Prove Inverse of Bijection function

  1. Jul 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose f is bijection. Prove that f⁻¹. is bijection.

    2. Relevant equations

    A bijection of a function occurs when f is one to one and onto.
    I think the proof would involve showing f⁻¹. is bijective, by showing f⁻¹ is onto, and one to one, since f is bijective it is invertible.

    3. The attempt at a solution

    To start:

    Since f is invertible/bijective
    f⁻¹ is one-to-one:

    f(a)=b then f⁻¹(b)=a
    if f(a)=b f(a)=b' then b=b'
    So f⁻¹ is one-to-one

    f⁻¹ is onto:
    *for f to be onto:
    ∀a∈A,∃b∈B f(a)=b
    ∀b∈B,∃a∈A f(b)=a
    So f⁻¹ is onto.

    Therefore f⁻¹ is bijective.
    Would this be correct?
    Thank You Very Much.
  2. jcsd
  3. Jul 6, 2009 #2


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    Science Advisor

    Yes, if f(a)= b and f(a)= b' then b= b'. But that is true of any function, not just one-to-one functions so it does NOT show that f-1 is one-to-one. You need to start from "if f-1(a)= f-1(b)" and finish "therefore a= b". I would recommend taking f of both sides.

  4. Jul 6, 2009 #3
  5. Jul 6, 2009 #4


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    Homework Helper

    I very much doubt that the book you use defines a surjection like that, but if it does it's simply wrong. The definition of a surjection is that [itex]\forall b \in B \;\; \exists a \in A , \text{such that} f(a)=b[/itex]. In words this means that a function's values span its whole codomain.

    f:[0,1] \rightarrow \{1,0\}, f(x)=1

    Using your definition of a surjection would allow you to conclude that this function, f, is a surjection. Seeing as for all values of [itex]x \in [0,1] [/itex] there is an [itex]y \in \{0,1\}[/itex] namely 1. Yet 0 never is reached for any [itex]x \in [0,1][/itex] thus it cannot be a surjection.
    Now lets use the correct definition of a surjection. For [itex]y=1[/itex] there is always an [itex]x \in [0,1][/itex] such that [itex]f(x)=1[/itex], namely all values within [0,1]. But for y=0 there is not a single value for [itex]x \in [0,1][/itex] such that f(x)=0. So there exist an [itex]y \in \{0,1\}[/itex] such that [itex]f(x) \neq y[/itex], namely y=0. Therefore this function is not a surjection.
    Last edited: Jul 6, 2009
  6. Jul 6, 2009 #5
    Oh my apologies I made a mistake, you are right, I am working on some inverse function stuff and typing functions gets me confused very easily. But thank you.
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