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Math Help - Is this composition bijective?

  1. #1
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    Is this composition bijective?

    If f:X->Y is bijective and g is any map s.t g:X->X

    then f \;o \; g \;o f^{-1} is bijective.



    the compostion is a function that goes from Y to Y. I am pretty sure it's onto, since f is onto. It's injective because g uniquely determines what the composition ends up as.

    (we are sending g:X-X to a unique function, that goes from Y->Y)

    I think the real question I'm asking is, does this define an isomorphism?

    Sorry if this is unclear. I tried my best to word it coherently. I will try to clarify if needed.
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  2. #2
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    Re: Is this composition bijective?

    Quote Originally Posted by Sheld View Post
    If f:X->Y is bijective and g is any map s.t g:X->X
    then f \;o \; g \;o f^{-1} is bijective.
    Let X=Y=\mathbb{R} and f(x)=x~\&~g(x)=x^2.

    f\circ g\circ f^{-1}(-1)=f\circ g\circ f^{-1}(1), bijective?
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  3. #3
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    Re: Is this composition bijective?

    suppose g(x1) = g(x2) for x1 ≠ x2 in X.

    since f is bijective, f^-1 is bijective, thus surjective.

    so we can find y1,y2 in Y with f^-1(y1) = x1, f^-1(y2) = x2.

    note that y1 ≠ y2 (since f^-1 exists, it is a function).

    now fogof^-1(y1) = f(g(f^-1(y1))) = f(g(x1)) = f(g(x2)) = f(g(f^-1(y2))) = fogof^-1(y2),

    so the composition is not injective, so it can't be bijective, either.
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