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

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

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

$\displaystyle [0, +\infty) = \{x \in \mathbb{R} : 0 \leq x < +\infty\}$,$\displaystyle (0, 1] = \{x \in \mathbb{R} : 0 < x \leq 1\}$

2. Originally Posted by pikminman
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 $\displaystyle f : [0, +\infty) \rightarrow (0, 1]$ be the function from the set $\displaystyle [0, +\infty)$ to the set
(0, 1] defined such that

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

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

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

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

EDIT: To fully illustrate this last point let $\displaystyle \phi:X\mapsto Y$ be bijective function. Define $\displaystyle \phi^{-1}(x)=\text{the unique pre-image of }x$. Then clearly
$\displaystyle \phi\circ\phi^{-1}=\iota_Y$ and $\displaystyle \phi^{-1}\circ \phi=\iota_X$, where $\displaystyle \iota_X$ is the identity mapping on $\displaystyle X$. Note that we needed bijectivity so that $\displaystyle \phi$ was well defined.

3. Originally Posted by Drexel28
Let $\displaystyle \kappa\in(0,1]$. Then $\displaystyle \frac{1}{\kappa}>1\implies \sqrt{\frac{1}{\kappa}}>1\implies \frac{1}{\sqrt{\kappa}}-1>0$. Therefore $\displaystyle \frac{1}{\sqrt{\kappa}}-1\in\text{Dom }f$ and $\displaystyle f\left(\frac{1}{\sqrt{\kappa}}-1\right)=\kappa$. Thus $\displaystyle f$ is surjective.

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

EDIT: To fully illustrate this last point let $\displaystyle \phi:X\mapsto Y$ be bijective function. Define $\displaystyle \phi^{-1}(x)=\text{the unique pre-image of }x$. Then clearly
$\displaystyle \phi\circ\phi^{-1}=\iota_Y$ and $\displaystyle \phi^{-1}\circ \phi=\iota_X$, where $\displaystyle \iota_X$ is the identity mapping on $\displaystyle X$. Note that we needed bijectivity so that $\displaystyle \phi$ was well defined.
Ahhh thank you! I don't understand the last part though.

4. Originally Posted by pikminman
Ahhh thank you! I don't understand the last part though.
If $\displaystyle \phi$ is bijective, then for any [tex]y\inY[/mah] there is exactly one $\displaystyle x\in X$ such that $\displaystyle \phi(x)=y$. So define $\displaystyle \phi^{-1}:Y\mapsto X$ as $\displaystyle \phi^{-1}(y)=x$. That is a function such that $\displaystyle \phi\circ\phi^{-1}=\phi^{-1}\circ\phi=x$. Which is the definition of an inverse.

5. Originally Posted by Drexel28
If $\displaystyle \phi$ is bijective, then for any $\displaystyle y \in Y$ there is exactly one $\displaystyle x\in X$ such that $\displaystyle \phi(x)=y$. So define $\displaystyle \phi^{-1}:Y\mapsto X$ as $\displaystyle \phi^{-1}(y)=x$. That is a function such that $\displaystyle \phi\circ\phi^{-1}=\phi^{-1}\circ\phi=x$. Which is the definition of an inverse.
Cool!