How do I verify if $\phi : \mathbb{Z}_{6} \rightarrow \mathbb{Z}_{36}$ is well-defined?

Question

Given $\phi : \mathbb{Z}_{6} \rightarrow \mathbb{Z}_{36}$ and $\phi ([a]_{6})=[a]_{36}$, verify that $\phi$ is a well-defined function.

My understanding is that well-defined is the converse of injective. So given a function $f:A\rightarrow B$, $f$ is well-defined when for $a = b$ in $A$, then $f(a) = f(b)$ in $B$.

Attempting to apply this definition, this is where I end up:

Say $[a]_6 = [b]_6$

$\Rightarrow 6$ | $(a - b)$

$\Rightarrow 36$ | $6(a - b)$

$\Rightarrow 36$ | $(6a - 6b)$

So $[6a]_{36} = [6b]_{36}$

Which is not $\phi([a]_{6}) = \phi([b]_{6})$

Does this show that $\phi$ is not a well-defined function? If not, where am I going wrong?

Answer 1

$\phi$ is not well defined.

$[6]_6 = [12]_6$ because $6 | 12 - 6$ but $[6]_{36} \ne [12]_{36}$ since $36 \nmid 12 - 6$.

Answer 2

Obviously $\phi$ is not well defined.

If two numbers are congruent modulo 6 it does not follow that they are also congruent modulo $36$.

For example $[5]=[11]$ mod($6$) but $[5]\ne [11]$ mod ($36$)

If we reverse the role of domain and codomain then we will have a well defined function.

But that is not the same function any more.

Answer 3

To prove that a statement is false, you just need to find a counter example. You cannot say "this statement is false because I cannot prove it". For here, you just need to find some $[a]_{6}=[b]_{6}$ such that $\phi([a]_{6})\neq \phi([b]_{6})$.