Proof the bijectivity of the exponential function mod $2 \pi i$

Question

I am trying to show that the map $\Psi\ \colon \left\{\begin{align}\mathbb{R}/\mathbb{Z} & \longrightarrow S^1 \\ x & \longmapsto e^{2\pi i x}\end{align}\right.$ is a bijection.

I am stuck on proving that it is injective.

I suppose that $\Psi(x)=\Psi(y) \iff e^{2 \pi i x}=e^{2 \pi i y} \iff x=y$ but I am struggling to see how to write this more formally as this seems way too sketchy.

Answer 1

$\Psi(x)=\Psi(y)$ implies that $e^{2\pi ix}=e^{2\pi iy}$ implies that $2\pi x=2\pi y+k\pi$ for some $k\in \Bbb Z$, implies $x-y=k\in \Bbb Z$, hence $x=y$ in $\Bbb R/ \Bbb Z$.

Answer 2

Since the map in question is a group homomorphism, it is enough to show that the kernel is trivial, i.e. for every $x\in\mathbb{R}$$$e^{2\pi i x}=1\Leftrightarrow x\in\mathbb{Z},$$and the latter follows immediately from a straightforward calculation.

Answer 3

For $x,y$ in $\mathbb{R}/\mathbb{Z} :$

$\Psi(x)=\Psi(y) \iff e^{2 \pi i x}=e^{2 \pi i y} \iff e^{2 \pi i (y-x)}= 1 \iff 2 \pi (y-x) = 0 \iff x=y$