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Math Help - Surjective function proof

  1. #1
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    Surjective function proof

    How do I prove that this is surjective? I already proved that it's injective.

    Also, if a function in both injective and surjective, is that function automatically invertible?

    Let f : [0, +\infty) \rightarrow (0, 1] be the function from the set [0, +\infty) to the set
    (0, 1] defined such that

    f(x) =\dfrac{1}{1 + x^2}

    for all x \in [0, +\infty), where

    [0, +\infty) = \{x \in \mathbb{R} : 0 \leq x < +\infty\}, (0, 1] = \{x \in \mathbb{R} : 0 < x \leq 1\}
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by pikminman View Post
    How do I prove that this is surjective? I already proved that it's injective.

    Also, if a function in both injective and surjective, is that function automatically invertible?

    Let f : [0, +\infty) \rightarrow (0, 1] be the function from the set [0, +\infty) to the set
    (0, 1] defined such that

    f(x) =\dfrac{1}{1 + x^2}

    for all x \in [0, +\infty), where

    [0, +\infty) = \{x \in \mathbb{R} : 0 \leq x < +\infty\}, (0, 1] = \{x \in \mathbb{R} : 0 < x \leq 1\}
    Let \kappa\in(0,1]. Then \frac{1}{\kappa}>1\implies \sqrt{\frac{1}{\kappa}}>1\implies \frac{1}{\sqrt{\kappa}}-1>0. Therefore \frac{1}{\sqrt{\kappa}}-1\in\text{Dom }f and f\left(\frac{1}{\sqrt{\kappa}}-1\right)=\kappa. Thus f is surjective.


    And yes, if a function is bijective then it is invertible.

    EDIT: To fully illustrate this last point let \phi:X\mapsto Y be bijective function. Define \phi^{-1}(x)=\text{the unique pre-image of }x. Then clearly
    \phi\circ\phi^{-1}=\iota_Y and \phi^{-1}\circ \phi=\iota_X, where \iota_X is the identity mapping on X. Note that we needed bijectivity so that \phi was well defined.
    Last edited by Drexel28; November 17th 2009 at 03:44 PM.
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    Quote Originally Posted by Drexel28 View Post
    Let \kappa\in(0,1]. Then \frac{1}{\kappa}>1\implies \sqrt{\frac{1}{\kappa}}>1\implies \frac{1}{\sqrt{\kappa}}-1>0. Therefore \frac{1}{\sqrt{\kappa}}-1\in\text{Dom }f and f\left(\frac{1}{\sqrt{\kappa}}-1\right)=\kappa. Thus f is surjective.


    And yes, if a function is bijective then it is invertible.

    EDIT: To fully illustrate this last point let \phi:X\mapsto Y be bijective function. Define \phi^{-1}(x)=\text{the unique pre-image of }x. Then clearly
    \phi\circ\phi^{-1}=\iota_Y and \phi^{-1}\circ \phi=\iota_X, where \iota_X is the identity mapping on X. Note that we needed bijectivity so that \phi was well defined.
    Ahhh thank you! I don't understand the last part though.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by pikminman View Post
    Ahhh thank you! I don't understand the last part though.
    If \phi is bijective, then for any [tex]y\inY[/mah] there is exactly one x\in X such that \phi(x)=y. So define \phi^{-1}:Y\mapsto X as \phi^{-1}(y)=x. That is a function such that \phi\circ\phi^{-1}=\phi^{-1}\circ\phi=x. Which is the definition of an inverse.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    If \phi is bijective, then for any y \in Y there is exactly one x\in X such that \phi(x)=y. So define \phi^{-1}:Y\mapsto X as \phi^{-1}(y)=x. That is a function such that \phi\circ\phi^{-1}=\phi^{-1}\circ\phi=x. Which is the definition of an inverse.
    Cool!
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