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F bijective <=> f has an inverse

  1. Sep 11, 2016 #1

    Math_QED

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    1. The problem statement, all variables and given/known data

    Proof that: f has an inverse ##\iff## f is a bijection

    2. Relevant equations/definitions

    A) ##f: X \rightarrow Y##

    If there is a function ##g: Y \rightarrow X## for which ##f \circ g = f(g(x)) = i_Y## and ##g \circ f = g(f(x)) = i_X##, then ##g## is the inverse function of ##f##.

    B) A function ##f## is a bijection if
    ##\forall y \in Y, \exists! x \in X: y = f(x)##
    or ##f## is surjective and injective
    or there is a function ##g: Y \rightarrow X## for which ##f \circ g = f(g(x)) = i_Y## and ##g \circ f = g(f(x)) = i_X##

    3. The attempt at a solution

    ##\Rightarrow##

    ##f##

    ##f: X \rightarrow Y## has an inverse, thus ##f \circ g = f(g(x)) = i_Y## and ##g \circ f = g(f(x)) = i_X##. We must show that f is bijective, so we will show it is both injective and surjective.

    1) Take ##x_1,x_2 \in X##

    ##f(x_1) = f(x_2)##
    ##\Rightarrow g(f(x_1)) = g(f(x_2))##
    ##\Rightarrow i_X(x_1) = i_X(x_2)##
    ##\Rightarrow x_1 = x_2##

    Thus, f is injective.

    2) Take ##y \in Y##

    Then, ##y = i_Y(y) = f(g(y))##.

    Since g is a function ##Y \rightarrow X##, we have ##\forall y \in Y, \exists g(y) \in X: y = f(g(y))## thus f is surjective.

    ##\Leftarrow##

    ##f## is a bijection, thus we have that there is a function ##g: Y \rightarrow X## for which ##f \circ g = f(g(x)) = i_Y## and ##g \circ f = g(f(x)) = i_X##. And this is exactly the definition of the inverse function.

    So, QED. Or not? Now comes the actual question. Is the definition my textbook provides for bijection wrong? I think this, because what I actually have to proof, is in the definition. I do understand that it is a useful criterium to see whether a function is a bijection, but should it be in the very definition of bijection? Thank you.
     
  2. jcsd
  3. Sep 11, 2016 #2
    No, the second part of the proof is incorrect. You have assumed the definition of bijective is equivalent to the definition of having an inverse, before proving it. Assume ##f## is a bijection, and use the definition that it is both surjective and injective. Then use surjectivity and injectivity to show some ##g## exists with the properties of the inverse.
     
  4. Sep 11, 2016 #3

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    So now I need to show that

    ##f## is a bijection ##\Rightarrow f## has in inverse.

    Proof: ##f## is a bijection. Thus it is both injective and surjective. This means, that:
    ##f(x_1) = f(x_2) \Rightarrow x_1 = x_2## and ##\forall y \in Y, \exists x \in X: y = f(x)##

    Now, we need to proof that there exists a function ##g: Y \rightarrow X## for which ##f \circ g = f(g(x)) = i_Y## and ##g \circ f = g(f(x)) = i_X##.
    So, let us first define ##g: Y \rightarrow X##:

    ##g(y) = x \iff f(x) = y##. (we can write this because f is a bijection, this is why the surjectivity is nessecary)

    Take ##x_1,x_2 \in X##: Then we can write:

    ##g(f(x_1)) = x_2 \Rightarrow f(x_2) = f(x_1) \Rightarrow x_2 = x_1##. Thus, ##g(f(x)) = x = i_X##
    Analogue, we obtain: ##f(g(x)) = x = i_Y##. Thus, ##g## is the inverse function of ##f## by definition.

    I'm not to sure about the way I defined ##g## though.
     
  5. Sep 11, 2016 #4

    PeroK

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    That's not so much wrong as not quite right! You can patch it up logically as follows:

    We define ##g: Y \rightarrow X##:

    For ##y \in Y##, we define ##g(y) = x## where ##f(x) = y##.

    ##g## is well-defined as ##\forall y \in Y, \exists## unique ##x## such that ##f(x) = y\ ## (as ##f## is a bijection).

    In general, what you have done is not wrong but often you have done more than is necessary, For example:

    If ##f## has an inverse, then ##f## is surjective:

    For ##y \in Y##, ##f(f^{-1}(y)) = y##.

    Hence ##f## is surjective.

    That's actually all you need.
     
  6. Sep 11, 2016 #5

    Math_QED

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    Thanks a lot for this.

    What have I done more here that was unnessecary? Can you point out other places where I did too much?
     
  7. Sep 11, 2016 #6

    PeroK

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    You could have missed out the bit I've marked **. Your logic stands up without spelling it out.

    I think it's a natural process when you learn this to say things like "as ##g## is a function". It's not wrong and - perhaps it's even a good thing at this stage. But, it's useful to know that you can start to drop that. And, if you have more difficult problems, you can soon get lost in all the detail and lose the wood in the trees.
     
  8. Sep 11, 2016 #7

    Math_QED

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    Is that the only thing that was unnessecary? I believe you are right. Those are things that may be obvious to you and other people who have a lot of experience in mathematics, but all the concepts are rather new for me and it only becomes obvious to me when I write it out in that way. Probably later..

    I hope the proof is correct now. Thanks a lot for your help. You have helped me out a lot of times.
     
  9. Sep 11, 2016 #8

    PeroK

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    The proof is well done. And, the more confident you become, the more you will want to drop some of the details yourself.
     
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