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Math Help - Every bijection has a two sided inverse

  1. #1
    Member oldguynewstudent's Avatar
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    Every bijection has a two sided inverse

    I believe I have the same professor G as mathgirl13 and the whole class is struggling.

    I need to prove the converse of the above:

    If f: X \longrightarrow Y has a two-sided inverse, prove that it is a bijection.

    Given (f^{-1}\circ f)(x) = x and (f^{-1}\circ f)(y) = y
    T.P. (f\circ f^{-1})(y1) = (f\circ f^{-1})(y2) \longrightarrow y1 = y2

    (f\circ f^{-1})(y1) = y1 = (f\circ f^{-1})(y2) = y2
    Now choose (f^{-1})(y) =x. This should prove f is injective, right?

    Next T.P.  \forall  y  \epsilon  Y  \exists  x  \epsilon  X  |  f(x) = y
    Choose y=f(x) then (f^{-1})(f(x)) =y

    but also (f^{-1})(f(x)) = f(x)  therefore f(x)=y \forall y \epsilon Y
    This should prove f is surjective, right?

    Thanks for any and all help.
    Last edited by oldguynewstudent; October 25th 2009 at 07:33 PM. Reason: math tags
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  2. #2
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    Quote Originally Posted by oldguynewstudent View Post
    I need to prove the converse of the above: If f: X \longrightarrow Y has a two-sided inverse, prove that it is a bijection.
    You want to show that f is a bijection.

    Suppose that f(a)=f(b) then you know that
    f^{-1}(f(a))=a~\&~f^{-1}(f(b))=b.
    Because f(a)=f(b) that means a=b.

    You seem to have done the surjective part.
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