1. ## 1:1 and onto

Is it possible for a function to be 1:1 and not onto and is it possible for a function to be onto but not 1:1? The catch is the function is from a set A->A. I can't find a set A and a function f so f:A->A that fits these (two separate functions and sets to meet each condition). Any hints on this are greatly appreciated.

2. Yes, it is possible.

3. Originally Posted by vassago
Is it possible for a function to be 1:1 and not onto and is it possible for a function to be onto but not 1:1? The catch is the function is from a set A->A. I can't find a set A and a function f so f:A->A that fits these (two separate functions and sets to meet each condition). Any hints on this are greatly appreciated.
Is it possible for a function to be 1:1 and not onto and is it possible for a function to be onto but not 1:1?
$\displaystyle f: \mathbb{Z} \rightarrow \mathbb{Z}$ by $\displaystyle x \mapsto 2x$. 1-1 not onto

a function to be onto but not 1:1?
$\displaystyle f: \mathbb{Z} \rightarrow \{0\}$ by $\displaystyle x \mapsto 0$. onto not 1-1

4. Originally Posted by GaloisTheory1
$\displaystyle f: \mathbb{Z} \rightarrow \mathbb{Z}$ by $\displaystyle x \mapsto 2x$. 1-1 not onto

$\displaystyle f: \mathbb{Z} \rightarrow \{0\}$ by $\displaystyle x \mapsto 0$. onto not 1-1
Of, if you must have functions that map a set onto itself, f: \mathbb{Z} \rightarrow \mathbb{Z}[/tex], f(x)= x if $\displaystyle x\le 0$, f(x)= x-1 for x> 0. That is onto but not 1-1 since f(0)= f(1)= 0.

If A were a finite set neither of these would be possible.