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Inverse function proof

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data

    If f:x-->y has an inverse function g, then g:y--->x is one to one and onto

    2. Relevant equations



    3. The attempt at a solution
    Let g be the inverse of f:x--->y
    I think this must have something to do with an isomorphism because of one to one and onto.
     
  2. jcsd
  3. Mar 30, 2009 #2
    has more to do with the function being bijective which is both one-one and onto (injective and surjective)

    just have to work the definitions, one-one means that only one distinct element gets mapped, and onto implies that the entire range has to be mapped, together these conditions guarantee that only one element gets uniquely mapped to another element throughout the domain we are working in.
     
  4. Mar 30, 2009 #3
    Ok, I understand those defintions. My problem is understanding why g has to be bijective.
     
  5. Mar 30, 2009 #4
    Do you know how to prove that a function is bijective?
     
  6. Mar 30, 2009 #5
    show a(x)=a(y) implies x=y
    onto, not entirely sure. I know it has something to do with showing entire range is used
     
  7. Mar 30, 2009 #6
    Not the range, but the codomain, i.e. for every y in the codomain, there is an x in the domain with y = f(x).
     
  8. Mar 30, 2009 #7
    Let g be the inverse of f:y--->x

    Then g(a)=g(b)
    a=b
    Does this prove one to one? For some resaon I think it doesn't.
     
  9. Mar 30, 2009 #8
    There is an intermediate step between "g(a) = g(b)" and "a = b". How would get rid of the g in g(a) to get a?
     
  10. Mar 30, 2009 #9
    I'm blanking on that
     
  11. Mar 30, 2009 #10

    HallsofIvy

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    Apply the inverse of g to both sides of the equation.
     
  12. Mar 30, 2009 #11
    so g^(-1)g(a)=g^(-1)g(b)?
     
  13. Mar 30, 2009 #12
    If g is the inverse of f, then what is the inverse of g? Also, what does g^{-1}(g(a)) evaluate to?
     
  14. Mar 30, 2009 #13
    f is the inverse of g?
     
  15. Mar 30, 2009 #14
    That's right. And what about the answer to my second question.
     
  16. Mar 30, 2009 #15
    f(g(a))
     
  17. Mar 30, 2009 #16
    And that simplifies to ...
     
  18. Mar 30, 2009 #17
    a I think
     
  19. Mar 31, 2009 #18
    That's right. Now what is the complete proof that g is one-to-one?
     
  20. Mar 31, 2009 #19
    by showing a=b
     
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