Let $f:S^1 \to S^1$ be given by $f(z)=z^2$, where $z=x+iy, x^2+y^2=1$. Then find the unique lift $\bar f: S^1 \to \mathbb{R}$ with the properties that (i) $\bar f(1)=0$ and (ii) $E \circ \bar f=f$, where $E:\mathbb{R} \to S^1$ is the map $E(t)=e^{2\pi it}=cos(2\pi t)+isin(2\pi t)$.
Please help, I couldn't find the lifting $\bar f$. Thanks a lot.
Recall De Moivre's formula that says that $(cos(\theta)+isin(\theta))^n=cos(n\theta)+isin(n\theta)$. So now it becomes clear that since elements of $S^1$ are only dependent on their arguments, the map $\bar{f}$ simply needs to send an element of $S^1$ to twice its argument which is an element of $\mathbb{R}$, (with some refinement with the way you have set up the map $E$, personally I would have described $E$ simply to be $t \mapsto e^{it}$). I am concerned that the map $\bar{f}$ will be discontinuous but this is the only option if you wish for $E \circ \bar{f}=f$.