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Thread: Every bijection has a two sided inverse

  1. #1
    Senior 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 $\displaystyle \longrightarrow $ Y has a two-sided inverse, prove that it is a bijection.

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

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

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

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

    Thanks for any and all help.
    Last edited by oldguynewstudent; Oct 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 $\displaystyle \longrightarrow $ Y has a two-sided inverse, prove that it is a bijection.
    You want to show that $\displaystyle f$ is a bijection.

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

    You seem to have done the surjective part.
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