why is this = [0],[1],[2],[3],[4],[5],[6]
and how would I define f Z/6Z - Z/6Z by f([a]) = ([2a]). I have the proof but I don't understand it.
Proof: Assume [a1] = [a2] in Z/6Z. then a1 congruent to a2(mod 6), so 6| a1 - a2. Thus, 6 | 2(a1 -a2) don't know why we use a 2 to multiply it all by property of divisibility, hence 6 | 2a1 - 2a2. But then 2a1 congruent to 2a2(mod 6) so by definition 2a1 =2a2 and finally f([a1]) = f([a2]).
Not sure how that conclusion is reached or what happened exactly.
The goal was to prove that $f$ is a well defined function, i.e. it assigns exactly one value to any element of its domain.
$\Bbb Z/6\Bbb Z$ contains six elements. Note that $[6]=[0]$ therein (and $[7]=[1]$ and $[8]=[2]$ and $[12]=[0]$ again, and $[-1]=[5]$, and so on..).
So, we take one element of this quotient set, which is one equivalence class. But we can only work by its representatives, the integers. So it starts with assuming that this class is represented by different integers $a_1$ and $a_2$. Then, we write the definition, multiply by $2$ because we want to prove that $f=[x]\mapsto[2x]$ is well defined, and arrive to $[2a_1]=[2a_2]$, which shows that the value of $f([x])$ is indeed independent of the choice of representative $x$ of the equivalence class.