The map $f:\mathbb{Z}_3 \to \mathbb{Z}_6$ given by $f(x + 3\mathbb{Z}) = x + 6\mathbb{Z}$ is not well-defined By naming an equivalence class in the domain that is assigned at least two different values prove that the following is not a well defined function.


  $$f : \Bbb Z_{3} \to \Bbb Z_{6} \;\;\;\text{ given by } f(\overline x) = [x]  $$
  
  In this case we represent an element of the domain as an $\bar x$ and use the notation $[x]$ for equivalence classes in the co-domain.


$f(\overline0) =  [0] \;,$  $ \Bbb Z_{3}  \quad (3x+0)\;\; \overline 0 = \{ ...-6,-3,0,3,6...  \}, \; \Bbb Z_{6}\;  (6x+0)\; \overline0 =\{ ...-12,-6,0,6,12...\}$ 

$f(\overline1) = [1], $ $\qquad   \; (3x+1) \; \;\;\;\overline   1 = \{ ...-5,-2,1,4,7 ... \},\; \;  (6x+1)\;\overline1 =\{...-11,-5,1,7,13.. \}$

$f(\overline2) = [2], $ $\qquad  \qquad \qquad \;\overline 2 = \{ ...-4,-3,2,5,8 ... \},\;\;\overline 2 = \{ ...-10,-4,2,8,14  ...\},\;$

$f(\overline3) = [3] ,$  $\qquad \qquad \qquad \qquad \qquad \qquad,\; \quad \quad \quad \; \; \; \;\overline 3 = \{ ...-9,-3,3,9,15 ... \},$

$f(\overline4) =  [4],\qquad \qquad \qquad\qquad \qquad\qquad \; \quad \quad \quad \quad \; \;\overline 4 = \{ ...-8,-2,4,10,16...  \}, $ 

$f(\overline5) = [5], \qquad \qquad \qquad\qquad \qquad\qquad \; \quad \quad \quad \quad \;\;\overline 5 = \{ ...-7,-1,5,11,17...  \},  $

$f(\overline6) = [6] ,$

So my main question for this problem is how to find out if this question is not a function. From the information I have gathered here I still cannot see why this is not a function any help on showing how this is not function would be much appreciated.


  The set of equivalence classes for the relation $\cong_{m}$ is denoted $\Bbb Z_{m}$


The $ 3x+0 \text{ and } 6x+0$ are just showing how I got $\overline 0 $
If a function $f \colon \mathbb{Z}_3 \to \mathbb{Z}_6$ with $f(\overline{x}) = [x]$ for every $x \in \mathbb{Z}$ would exist, then
$$
    [0]
  = f(\overline{0})
  = f(\overline{3})
  = [3],
$$
where we used for the second equality that $\overline{0} = \overline{3}$.
By the definition of $\mathbb{Z}_6$ we have for all $x, y \in \mathbb{Z}$ that $[x] = [y]$ if and only if $x - y$ is divisible by $6$, i.e. if $x - y \in 6\mathbb{Z} = \{6n \mid n \in \mathbb{Z}\}$.
So $[0] = [3]$ is equivalent to $3 - 0 = 3$ being divisible by $6$, which doesn’t hold.

This contradiction shows that no such function $f$ exists.

More generally one can show that there exists a function $g \colon \mathbb{Z}_n \to \mathbb{Z}_m$ with $g(\overline{x}) = [x]$ for every $x \in \mathbb{Z}$ if and only if $m$ divides $n$.
It is an onto relation from a set of lesser cardinality to a set of larger cardinality. Thus it cannot be a function (by the definition of cardinality).