# Thread: Surjective function proof

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 $f : [0, +\infty) \rightarrow (0, 1]$ 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

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

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 $f : [0, +\infty) \rightarrow (0, 1]$ 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.

3. Originally Posted by Drexel28
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.

4. Originally Posted by pikminman
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.

5. Originally Posted by Drexel28
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!