Let $\sim$ be an equivalence relation on $\mathbb{R}$ defined by
$$\forall x,y \in \mathbb{R}, x \sim y \iff x - y \in \mathbb{Z}$$
Let $\mathbb{U}$ be the set of all complex numbers of modulus 1, I need to show that the map $\dot{g}: \mathbb{R}/{\sim} \to \mathbb{U}$ is bijective. We have
$$\mathbb{R} \overset{s}{\longrightarrow} \mathbb{R}/{\sim} \overset{\dot{g}}{\longrightarrow} \mathbb{U}$$
We must construct a map $f: \mathbb{R} \to \mathbb{U}$ and $f = s \circ \dot{g}$ associated with the equivalence relation $\sim$ such that
$$\forall x,y \in \mathbb{R}, x \sim y \iff f(x) = f(y)$$
We notice that the equivalence relation is satisfied if two numbers have the same decimal part such that
$$\forall x,y \in \mathbb{R}, x \sim y \iff (x - \lfloor x \rfloor) = (y - \lfloor y \rfloor)$$
So the quotient set $\mathbb{R}/{\sim}$ can be related with ${\left[\right.}0,1{\left[\right.}$ and the map $f$ associated with $\sim$ can be defined as
$$f(x) = \exp(i (x - \lfloor x \rfloor)) \implies \forall x \in \mathbb{R}, \lvert f(x) \rvert = 1$$
I wonder if my definition for $f$ is correct (since it verifies $x \sim y$ and $\lvert f(x) \rvert = 1$), my course defines $f$ as $x \mapsto \exp(2\pi ix)$ and I do not understand how this definition is derived.
We can rewrite $x \mapsto \exp(2\pi ix)$ as
$$\exp(2\pi ix) = \exp(i2\pi(\lfloor x \rfloor + (x - \lfloor x \rfloor))) = \exp(i(2\pi \cdot \lfloor x \rfloor + 2\pi (x - \lfloor x \rfloor)))$$
And since $\lfloor x \rfloor \in \mathbb{N}$ we can write $x \mapsto \exp(2\pi i (x - \lfloor x \rfloor))$ but we notice
$$\exp(i (x - \lfloor x \rfloor)) \neq \exp(2\pi i (x - \lfloor x \rfloor))$$
There is an infinite number of morphisms $f: \mathbb{R} \to \mathbb{U}$ as long as two real numbers with the same decimal part belongs to the same equivalence class, so $f$ must satisfies
$$\forall x \in \dot{x}, f(x) = \dot{g}(\dot{x})$$
As an example, we must have $f(12.5) = f(0.5)$ since the equivalence relation is true if two real numbers have the same decimal part. If the morphism $f$ is invariant under $\sim$ then we have an injective morphism $\dot{g}: \mathbb{R}/{\sim} \to \mathbb{U}$ such that
$$\mathbb{R} \overset{\sigma}{\longrightarrow} \mathbb{R}/{\sim} \overset{\dot{g}}{\longrightarrow} \mathbb{U}$$
We can construct a surjective morphism $\dot{f}$ by a restriction to the codomain of $\dot{g}$ to $f(\mathbb{R})$ so that we have $\dot{f}: \mathbb{R}/{\sim} \to f(\mathbb{R})$ such that
$$\mathbb{R} \overset{\sigma}{\longrightarrow} \mathbb{R}/{\sim} \overset{\dot{f}}{\longrightarrow} f(\mathbb{R}) \overset{\iota}{\longrightarrow} \mathbb{U}$$
It follows that $\dot{f}$ is also injective since it is derived from $\dot{g}$ and that $\iota: f(\mathbb{R}) \to \mathbb{U}$ is injective and defined as the identity map with the domain restriction from $\mathbb{U}$ such that
$$\forall z \in f(\mathbb{R}), \iota(z) = z$$
If we prove that $f(\mathbb{R}) = \mathbb{U}$ then $\dot{g} = \dot{f}$ and $\dot{g}$ is bijective. Because we know two real numbers with the same decimal part belongs to the same equivalence class we have a partition for $\mathbb{R}$
$$\mathbb{R} = \bigcup_{x \in \mathbb{R}} \dot{x}$$
The set of all complex numbers of modulus 1 is given by the Euler formula $\exp(i\theta) = \cos(\theta) + i\sin(\theta)$ so we have the morphism $f: \mathbb{R} \to \mathbb{U}$ with $f(x) = \exp(ix)$ and recall the equivalence relation $\sim$ is defined as
$$\forall x,y \in \mathbb{R}, x \sim y \iff x - y \in \mathbb{Z}$$
And that the function $f$ needs to be invariant under $\sim$ such that
$$\forall x,y \in \mathbb{R}, x \sim y \iff f(x) = f(y)$$
Two numbers of the same equivalence class must have the same output by $f$ and the equivalence relation is true if and only if two numbers have the same decimal part, so two numbers with the same decimal part must have the same output by $f$ as such
$$\forall x,y \in \mathbb{R}, x \sim y \iff f(x - \lfloor x \rfloor) = f(y - \lfloor y \rfloor)$$
With $f(x) = \exp(i(x - \lfloor x \rfloor))$ we do not have all the complex numbers of modulus 1, indeed, the set of all complex numbers of modulus 1 corresponds to all the complex numbers located on the unit circle given by
$$\mathbb{U} = \{\exp(i\theta) \ \vert \ 0 \leqslant \theta \leqslant 2\pi\}$$
and because $0 \leqslant x - \lfloor x \rfloor < 1$ we only have the complex numbers for less than one quarter around the unit circle. However we notice that
$$\exp(0) = \exp(2\pi i) = 1$$
and since $\exp(i\theta)$ is periodic of period $2\pi$, we can multiply by $2\pi$ to have the whole set of complex numbers around the unit circle with
$$\left(e^{i(x - \lfloor x \rfloor)}\right)^{2\pi} = e^{2\pi i(x - \lfloor x \rfloor)} = e^{i(2\pi x - 2\pi \lfloor x \rfloor)} = e^{2\pi ix}$$
We can then simplify the morphism $f: \mathbb{R} \to \mathbb{U}$ invariant under $\sim$ with $f(x) = \exp(2\pi ix)$ and the last step is to show that $f$ is surjective. We have
$$\forall z \in \mathbb{U}, \exists x \in \mathbb{R}, f(x) = z$$
Starting with $z = f(x)$ we have
$$z = \exp(2\pi ix) \iff \ln(z) = 2\pi ix \iff \frac{\ln(z)}{2\pi i } = x$$
For the set of all complex numbers of modulus 1 we have
$$\lvert z \rvert = 1 \implies \ln(z) = \Im(\ln(z)) \ \text{and} \ \forall a,b \in \mathbb{I}, \frac{a}{b} \in \mathbb{R}$$
We have $x \in \mathbb{R}$ and $f$ surjective so $\mathbb{U} \subset f(\mathbb{R})$ and the immediate consequence is that $\dot{g} = \dot{f}$ which means $\dot{g}$ is bijective. End of story.