is $f: \Bbb Z_3 → \Bbb Z_9, f(\bar x) \ =[2x]$ a function?

Question

I am confused about going from $\Bbb Z_3$ to $\Bbb Z_9$ and how that impacts the function. I am not sure where to start.

So far I have:

$$\begin{align} f(\overline0) &= [0], \\ f(\overline1) &= [2], \\ f(\overline2) &= [4], \\ f(\overline3) &= [6]. \end{align}$$

I'm not sure how to prove whether or not it is a function.

Answer 1

Since $\overline{0}=\overline{3}$ in your notation, we should get

$$f(\overline{0})=f(\overline{3})\tag{1}$$

(and so on) for the function $f$ to be well-defined (that is, the application of $f$ is independent of the choice of equivalence class representatives). However, the LHS of $(1)$ is $[0]$ in your notation, whereas the RHS is $[6]$, which is not $[0]$. This is because $9\nmid 0-6$.