Let the function $f:\mathbb{R}/\mathbb{Z}\to U_1$ be defined by $f(\bar{z})=e^{2\pi i z}$, where $U_1=\{z\in\mathbb{C}^*:|z|=1\}$. Prove that this funcstion is well-defined and a that it is group-homomorphism.
Well-defined:
Suppose $\bar{a},\bar{b}\in\mathbb{R}/\mathbb{Z}$ and $\bar{a}=\bar{b}$, so that $\bar{a},\bar{b}\in[0,1)$. Without loss of generality, we can suppose that $a>b$ which implies that $a$ can be written as $a=b+k, k\in\mathbb{Z}$.
Then we can see that:
$f(\bar{b})=e^{2\pi ib}=e^{2\pi i(a+k)}=e^{i(2\pi a+k2\pi)}=$ $\cos(2\pi a + k2\pi)+i\sin(2\pi a + k2\pi) =\cos(2\pi a)+i\sin(2\pi a)=$ $e^{2\pi ia}=f(\bar{a})$
We now have $\bar{a}=\bar{b}\implies f(\bar{a})=f(\bar{b})$, so $f$ is well-defined.
Group-homomorphism:
Suppose $\bar{a},\bar{b}\in\mathbb{R}/\mathbb{Z}$, then:
$f(\bar{a}+\bar{b})=f(\overline{a+b})=e^{2\pi i(a+b)}=e^{2\pi ia}*e^{2\pi ib}=f(\bar{a})*f(\bar{b})$, so $f$ is a group-homomorphism.
I'm wondering whether my proof is correct. I was thinking this might be too simple or missing some crucial steps. I'm also not sure about the operations for the groups, I guessed that of $U_1$ would be multiplication as their elements come from $\mathbb{C}^*$, but the homomorphism only made sense if the operation on $\mathbb{R}/\mathbb{Z}$ would be addition. Any tips?