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

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 $\phi([a]_{6})\neq \phi([b]_{6})$.
