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Proof of bijection of a function

  1. Apr 3, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider a bijection f = (A,B,F)
    Show that f^(-1) (inverse of f) is a bijection from B to A and that for any element x of A we have
    f^(-1)(f(x))=x

    3. The attempt at a solution

    For this proof can I use contradiction and the say f^(-1) is not a bijection from B to A or there exists an element x of A that f^(-1)(f(x)) != x

    If this is what I am supposed to prove. How would I go about doing it? If not what proof method should I use?
     
  2. jcsd
  3. Apr 3, 2014 #2

    jbunniii

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    To show that ##f^{-1}## is a bijection, you need to show three things:
    (1) ##f^{-1}## is a function. This is not necessarily true for a general ##f##. For example, if ##f : \mathbb{R} \rightarrow \mathbb{R}## is defined by ##f(x) = x^2##, and ##y > 0##, then ##f^{-1}(y)## is a set containing two distinct elements: ##\sqrt{y}## and ##-\sqrt{y}##.
    (2) ##f^{-1}## is an injection, meaning that if ##f^{-1}(y_1) = f^{-1}(y_2)##, then ##y_1 = y_2##.
    (3) ##f^{-1}## is a surjection, meaning that if ##x## is in the codomain of ##f^{-1}##, then ##x = f^{-1}(y)## for some ##y## in the domain of ##f^{-1}##.
     
  4. Apr 3, 2014 #3
    alright I got it.
     
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