What does it mean for f([x])=[2x] for a function mapping R/Z to R/Z? Let X=R/Z (the circle), with a map $f : X → X$ given by $f([x]) = [2x]$.
I'm a little lost on what $f([x]) = [2x]$ means. I thought the function was mapping the equivalence class [x] to the equivalence class double of x, but then I'm asked
What are the two elements of $f^{-1}([3/4])$? Verify that f is 2-to-1, 
meaning $|f^{−1}([x])| = 2$ for every $[x] ∈ X$. 
I thought that it would 3/8 or any variation like 6/16 for the first part, but I don't see how $|f^{−1}([x])| = 2$ for every $[x] ∈ X$.
 A: You are correct that $f([x])=[2x]$ means that $f$ maps the equivalence class of $x$ to the equivalence class of $2x$. So for example, the equivalence class of $\frac{1}{2}$ would be mapped to the equivalence class of $1$, which are obviously two different equivalence classes.
Now $f([x])=f([y])$ if and only if $[2x]=[2y]$, which is true if and only if $2x-2y=k\in\mathbb{Z}$. So $x=\frac{k}{2}+y$. Thus looking at $f^{-1}[0]$, i.e. when $y=0$, we can see that it only contains the classes $[\frac{1}{2}]$ and $[1]$. Can you generalize using this idea that $f^{-1}[y]$ always contains exactly two classes?
A: Suppose, $[x]\in \mathbb R/\mathbb Z$. We may assume, $0\le x<1$. Now, $[2y]=[x]$ if and only if $2y-x\in\mathbb Z$. In particular, for $x=0$, we must have $2y\in\mathbb Z$. So, $f^{-1}([0])=\left\{\left[\frac n2\right]:n\in\mathbb Z\right\}=\left\{[0],\left[\frac12\right]\right\}$.
If $x\ne 0$, then $f^{-1}([x])=\left\{\left[\frac x2+\frac n2\right]:n\in\mathbb Z\right\}=\left\{\left[\frac x2\right],\left[\frac{x+1}2\right]\right\}$.
