Describe as explicitly as you can all terms in the canonical decomposition of the function $\mathbb R \to \mathbb C$ as $r \mapsto e^{2\pi ir}$ $f:\mathbb R \to \mathbb R/ {\sim}$ as $r \mapsto [r]_\sim$ where $\sim$ on $\mathbb R$ is given by $r' \sim r'' \leftrightarrow f(r') = f(r'').$
$\widetilde f: \mathbb R/{\sim} \to \operatorname{im} f$ given as $\widetilde f([r]_\sim) = f(r).$
$f: \operatorname{im}f \to \mathbb C$ given as $[r]_\sim \mapsto e^{2\pi ir}$.
Is that what we are being asked to do?
 A: $$
\mathbb R \longrightarrow \hspace{-14pt} \longrightarrow (\mathbb R/{\sim}) \overset \sim \longrightarrow (\operatorname{im} f = \{ z\in \mathbb C : |z|=1\}) \hookrightarrow \mathbb C
$$
Since $r\mapsto e^{2\pi r i}$ is periodic with period $1$, the relation $a\sim b$ on $\mathbb R$ means $a-b$ is an integer.  You can show that that is an equivalence relation.
The space $\mathbb R/{\sim}$ is the set of all equivalence classes.  Then $1/8$ (think of a $45^\circ$ angle) becomes the same as $9/8$ (a $360^\circ+45^\circ = 405^\circ$ angle). The mapping from $\mathbb R$ onto $\mathbb R/{\sim}$ takes a number like $9/8$ and maps it to the equivalence class it belongs to, in effect forgetting whether it's $1/8$ or $9/8$ or $17/8$ or some other point where the sine and cosine are the same as they are at that point. I wrote "onto" rather than "into" because no equivalence class is omitted.
The mapping from $\mathbb R/{\sim}$ to $\operatorname{im} f$ goes from $\mathbb R/{\sim}$ the circle $\{z\in\mathbb C: |z|=1\}$.  That circle is $\operatorname{im} f$.  It takes the equivalence class to which $1/12$ belongs (and so does $1+1/12$, and so does $2+1/12$, and so does $-1+1/12$, etc) to the point
$$
e^{2 \pi i \left( \frac{1}{12} \right)} = \cos\frac\pi 6 + i\sin\frac\pi6 = \frac{\sqrt3}2 + i\frac 1 2.
$$
That point is on the circle $|z|=1$, which is the image of $f$.  This mapping is both one-to-one and onto.  It is one-to-one because it always takes different equivalence classes of numbers to different points on the circle.  It is onto because every point on the circle is the image of some equivance class.
Finally, the mapping from $\operatorname{im} f = \{ z\in\mathbb C: |z|=1\}$ to $\mathbb C$ just takes each point to itself.  It is one-to-one.  It is not onto because not every point in $\mathbb C$ is the image of some point on the circle.  Only those points in $\mathbb C$ that are on the circle have pre-images under this function.
