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 $

It is easy to show that
$\mathbb Z_3=\left\{\begin{array}\{
\{...,-6,-3,0,3,6,...\},\\
\{...,-5,-2,1,4,7,...\},\\
\{...,-4,-1,2,5,8,...\}\end{array}\right\}$
and that
$\mathbb Z_6=\left\{\begin{array}\{
\{...,-12,-6,0,6,12,...\},\\
\{...,-11,-5,1,7,13,...\},\\
\{...,-10,-4,2,8,14,...\},\\
\{...,-9,-3,3,9,15,...\},\\
\{...,-8,-2,4,10,16,...\},\\
\{...,-7,-1,5,11,17,...\}\end{array}\right\}$
, because $\mathbb Z_3$ and $\mathbb Z_3$ are sets of equivalence classes. Let the relation $f$ be represented by a diagram:

Here, the relation lines show that $f(\bar1)=[1]$, $f(\bar2)=[2]$, and $f(\bar3)=[3]$.
But consider that $\bar0\cong_3\bar3$, which means they are the same element in $\mathbb Z_3$ (Remember, $\mathbb Z_3$ is a set of equivalence classes). Then, consider that $f(\bar0)=f(\bar3)=[3]$. Putting this into the diagram yields

, which shows that $f$ is clearly not a function, because a function maps each element in its domain to exactly $1$ element in its co-domain. $Q.E.D.$
