Explain why the following statement is false

Question

Let $f:S^1 \to S^1$ be given by $f(z)=z^2$, where $z=x+iy, x^2+y^2=1$. Then there is a 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 $E(t)=e^{2\pi it}=\cos(2\pi t)+i\sin(2\pi t)$.

I thought this statement is true and I have proved it, which is completely worng proof, as I just been told by a professor that is a false statement. Can anyone help me to explain why it is a false statement? Thanks a lot!

Answer 1

$f$ induces a homomorphism $f_*$ where $f_*(\pi_1(S^1)) \cong 2\mathbb{Z}$. However, supposing you can lift $f$ to $\overline{f}$ then $\overline{f}_*$ induces a homomorphism into the trivial group $\pi_1(\mathbb{R})$. Note that $E(t) = e^{2\pi it}$ is a covering map for $S^1$ and induces a homomorphism $E_*:\pi_1(\mathbb{R}) \to \pi_1(S^1) \cong \mathbb{Z}$. Therefore $f_* \neq E_*\circ \overline{f}_*$.

Answer 2

The degree of the map $z \rightarrow z^2$ is 2.But if it factor through $R$ then the map is homotopically trival and so degree is zero which is a contradiction.