Lifting a homomorphism

Question

In a set of lecture notes, I recently came across the remark that any continuous homomorphism from $\mathbb{R} \to S^1$ can be lifted to continuous homomorphism from $\mathbb{R} \to \mathbb{R}$. How do I conclude that the lift is a homomorphism too? I tried many ways to solve this but failed. I could conclude, for example, that if the lift is $\bar{f}$, then for any $x_1, x_2 \in \mathbb{R}$, $\bar{f}(x_1+x_2)=\bar{f}(x_1)+\bar{f}(x_2)+2n\pi$ for some $n\in \mathbb{Z}$. If this $n$ were fixed, I might be able to say something, however this may not be true. I feel that this must be a very trivial problem, but I have spent a lot of time on it, without a result.

Answer 1

Fix $x_1$ and vary $x_2$ a little, then $\bar f(x_1+x_2)$ and $\bar f(x_2)$ change a little, while $\bar f(x_1)$ stays fixed. Hence $2n\pi$ also can only change a little. Since $\mathbb Z$ is discrete $n$ has in fact to be constant. Thus $n$ is constant on connected components of $\mathbb R$. Since $\mathbb R$ is connected you are done.