This may be really elementary but I'm lost. Let $S^1\subseteq\mathbb{C}$ be the unit circle, which is a group with complex multiplication. Let $\mathbb{T}:=\frac {\mathbb{R}}{\mathbb{Z}}$ be the $1$-torus, which is also a group with addition $\mod 1$. We call ($\textbf{unitary}$) $\textbf{character of}\; \mathbb{T}$ any continuous (with respect to the induced subspace topology) homomorphism $\phi:\mathbb{T}\rightarrow S^1$.
For example, for each $n\in\mathbb{Z}$, we have the following character
$$\chi_n(x):=e^{2\pi in x};$$
my aim is to prove that any character of $\mathbb{T}$ is of this form. In other words, given a character $\phi$ I want to find an integer $n$ such that $\phi=\chi_n$. In order to do so I was thinking to use the covering map of $S^1$ given by
$$\exp:\mathbb{R}\rightarrow S^1,$$
$$t\mapsto e^{2\pi i t},$$
which is a (continuous) group homomorphism with $\ker\exp=\mathbb{Z}$ (indeed, this shows that $\mathbb{T}\simeq S^1$ as groups). Now, I would like to take a character $\phi$ and lift it to a map $\phi^*:\mathbb{R}\rightarrow S^1$ which should be easier to handle, but this is where I get lost and cannot fill in the details. Any idea on how to proceed? If you have a completely different route I would also be glad to see that.
I don't know if you gain anything particular from lifting the map. Let $\phi$ be a character on $\mathbb{R}/\mathbb{Z}$ and let us analyse it. Throughout, I'll identify $\mathbb{R}/\mathbb{Z}$ with $[0,1)$ (i.e. a neighbourhood around $0$ has the form $[0,\varepsilon)\cup (1-\varepsilon',1)$).
The first thing to note is that $\phi$, being a homomorphism, necessarily reduces the order of elements. In particular it maps finite order elements to finite order elements and we get for the special points $t=\frac{1}{q}$ with $q\in \mathbb{N}$ that $$ \phi\left(\frac{1}{q}\right)=\exp\left(i\frac{p'}{q'}2\pi\right), $$ where $gcd(p',q')=1$, $q'>p'$ and $q'|q$. If you don't know that elements of this form are exactly the elements of $S^1$ of finite order, this is a good exercise. Hence, $\phi(1/q)=\chi_{n_q}(1/q),$ where $n_q=\frac{p' q}{q'}$.
The second thing to notice is the consistency requirement $n_{q}=n_{2q}$. Indeed, $$ \exp\left(i \frac{n_q}{q}2\pi\right)=\phi\left(\frac{1}{q}\right)=\phi\left(\frac{1}{2q}\right)^2=\exp\left(i \frac{n_{2q}}{2q} 2\pi\right)^2 $$
Now we're basically done because we've determined that $\phi$ is identical to $\chi_{n_2}$ on the set $\{2^{-n}|n\in \mathbb{N}\}$. It follows immediately from the homomorphism property that $\phi$ is identical to $\chi_{n_2}$ on all of $\{\frac{k-1}{2^{n}}|n,k\in \mathbb{N}, 0<k\leq 2^n\}$, which is a dense subset of $[0,1)$. Hence, by continuity, $\phi$ must agree with $\chi_{n_2}$ everywhere.