Could someone help correct, if necessary, my attempt at proving the following question ?
Let $\Bbb{Z}_3=\{[0]_3,[1]_3,[2]_3\}$ and $\Bbb{Z}_6=\{[0]_6,[1]_6,[2]_6,[3]_6,[4]_6,[5]_6\}$ be partitions of $\Bbb{Z}$ induced by congruence modulo 3 and 6 respectively.
Let the purported function $f:\Bbb{Z}_3 \rightarrow \Bbb{Z}_6$ be defined by $f([a]_3)=[2a]_6$ for all $[a]_3 \in \Bbb{Z}_3$.
Prove that $f$ is well defined.
Attempt:
Let the purported function $f:\Bbb{Z}_3 \rightarrow \Bbb{Z}_6$ be defined by $f([a]_3)=[2a]_6$ for all $[a]_3 \in \Bbb{Z}_3$.
The rule informally describes the subset $f$ of $\Bbb{Z}_6$ x $\Bbb{Z}_3$ defined by:
$(x,y) \in f \iff (\exists a\in\Bbb{Z}) (x=[a]_3 \text{ and } y=[2a]_6)$.
By defintion a well-defined function must satisfy the following two properties:
(i) For each $x \in \Bbb{Z}_3$ there is $y \in \Bbb{Z}_6$ such that $(x,y) \in f$.
(ii) if $(x,y) \in f$ and $(x,z) \in f$ then $y=z$.
Regarding (i):
Let $x \in \Bbb{Z}_3$.There is an integer $a$ with $x=[a]_3$.Consider the following corollary:
Let $m \ge 1$ be an integer.For each $k \in \Bbb{Z}$ there is an integer r in the list $0,\dots,m-1$ such that $[k]_m=[r]_m$
This suggest that for $b \in \Bbb{Z},b=2a$, there exists an integer $r$ in the list $0,\dots,5$ such that $[b]_6=[2a]_6=[r]_6$.
Therefore $([a]_3,[2a]_6) \in f$ as required.
Regarding (ii)
Assume $([a]_3,[b]_6) \in f$ and $([a]_3,[c]_6) \in f$. The objective is to prove that $[b]_6=[c]_6$.
Since $f([a]_3)=[2a]_6$ by defintion of the fiunction and $f([a]_3)=[b]_6$ by assumption, it follows by transitivity that $[2a]_6=[b]_6$.
Similarly,with $f([a]_3)=[2a]_6$ by defintion of the fiunction and $f([a]_3)=[c]_6$ by assumption, it follows by transitivity that $[2a]_6=[c]_6$.
By previously established theorems :
$b\equiv 2a\pmod 6 \iff 6|(b-2a) \iff b-2a=6k$ for some $k \in \Bbb{Z}$.
$c\equiv 2a\pmod 6 \iff 6|(c-2a) \iff c-2a=6g$ for some $g \in \Bbb{Z}$.
Then, with $2a=c-6g$ and $b-2a=6k$ it follows that $b-(c-6g)=6k$ and finally $b-c=6(k-g)$ and hence: $6|(b-c) \iff b\equiv c\pmod 6 \iff [b]_6=[c]_6$ as required.
Since the putative function $f$ satisfies both properties required for it to be well-defined,this concludes the proof.
Is this right? somehow (i) feels a bit iffy..(ii) feels right.
Thank you for your time in any case.