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 $
 A: 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 $n$.
A: You have a rule that given an element of $\mathbb{Z}_3$, you can write down an element of $\mathbb{Z}_6$.  The problem is that your rule is not well-defined.  To be well-defined means that even if you have two different ways to describe the same input, you get the same output.  
To be more explicit, observe that in $\mathbb{Z}_3$, the elements $\overline{0}$ and $\overline{3}$ are the same object even though they look different (they are defined by different numbers).  Because these are the same element, they should go to the same place under the map $f$.  Unfortunately, $f(\overline{0})=[0]$ and $f(\overline{3})=[3]$ and $[0]\not=[3]$ in $\mathbb{Z}_6$.  Therefore $f$ can't be a function!
The formal definition of well-defined is: 


*

*"If $a=b$, then $f(a)=f(b)$."  (Technically, this is not a good definition because writing $f(\cdot)$ implies that $f$ is a function).  


If we write functions in terms of relations, well-defined can be expressed as 


*

*"If $(a,b),(a,c)\in f$, then $b=c$.

A: 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. $Q.E.D.$
A: $\mathbb Z_3 = \{\overline{0},\overline{1}, \overline{3}\}$
where $\overline{0} = \{....,0,3,6,9....\}$
$\overline{1} = \{.....,1,4,7,10.....\}$
$\overline{2} = \{......,2,5,8,11....\}$
$\overline{3} = \overline{0}; \overline{4} = \overline{1};\overline{5} = \overline{2}$.
So if $f$ is well defined it'd have to be that  $f(\overline {0}) = [0]$ and $f(\overline{0}) = f(\overline{3}) = [3]$ so $[0] = [3]$.
But $\mathbb 6 = \{[0],[1],[2],[3],[4],[5]\}$
where $[0] = \{....0,6,12,18\}$
and $[1] = \{.....,1,7,13, 19\}$
etc.
Note $[0] = \{....0,6,12...\} \ne \{.....3,9,15...\} = [3]$.  But $\overline{0} = \overline{3}$
So $f(\overline{0}= [0]$ if we  write $\overline{0}$ as $\overline{0}$.  But $f(\overline{0} = [3]$ if we write $\overline{0}$ as $\overline{3}$.
So $f$ gives different and unequal output for different ways we interpret equal input.  So $f$ is poorly defined as it doesn't consistantly give equal output for equal input.
A: In your consideration of the equivalence classes of $0$, you looked at the classes in both $\mathbb{Z}_3$ and $\mathbb{Z}_6$, but for the rest you only looked at the classes in $\mathbb{Z}_6$. If you look at both, the answer will become clear: in $\mathbb{Z}_3$, $\overline{0}$ and $\overline{3}$ are both $\{\ldots, -6, -3, 0, 3, 6, \ldots\}$. So this single equivalence class is getting sent to both $\overline{0}$ and $\overline{3}$ in $\mathbb{Z}_6$, which are different things in $\mathbb{Z}_6$!
A: A function $f : \mathbb{Z}_3 \to X$ is just a function $\tilde f: \mathbb{Z} \to X$ making the following diagram commute:
\begin{matrix}
\mathbb{Z} & \xrightarrow{\tilde f} & X \\
\downarrow & & \parallel \\
\mathbb{Z}_3 & \xrightarrow{f} & X
\end{matrix}
where the left map is the projection $\mathbb{Z} \to \mathbb{Z}/3\mathbb{Z} = \mathbb{Z}_3$. Explicitly, a function $f:\mathbb{Z}_3 \to X$ is equivalent to a function $f:\mathbb{Z} \to X$ with $\tilde f(x + 3n) = \tilde f(x)$ for all $x\in \mathbb{Z}_3 = \mathbb{Z}/3\mathbb{Z}$ and all $3n\in 3\mathbb{Z}$. In this particular problem, you're trying to take $\tilde f$ to be the projection $\mathbb{Z} \to \mathbb{Z}/6\mathbb{Z} = \mathbb{Z}_6$, which does not satisfy $\tilde f(x) = \tilde f(x + 3)$.
A: 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).
