TITLE: Lift of $z^2$ map $S^1 \to S^1$
QUESTION [1 upvotes]: $\newcommand{\id}{\operatorname{id}}$It can be proved that identity map $\id: S^1 \to S^1$ does not lift to $\widetilde{\id} : S^1 \to \mathbb{R}$ such that $e^{\widetilde{\id(z)}i} = z$.
The standard proof is based on injectivity contradiction (the lift can not be bijective because otherwise it will be a homeomorphism and it is not injective than for some $z_0, z_1 \in S^1$ we have $z_0 = e^{\widetilde{\id(z_0)}i} = e^{\widetilde{\id(z_1)}i} = z_1$. Contradiction).
It seems like this contraction doesn't work for, say, $z \mapsto z^2$. So does this map lift to $S^1 \to \mathbb{R}$?

REPLY [1 votes]: Fact(general lifting lemma)  $p : (E, e_0) \to (B, b_0)$ be a covering map and $f : (X, x_0) \to (B, b_0)$ be a continuous map. $f$ lifts to $\widetilde{f} : (X, x_0) \to (E, e_0)$ (such that the obvious diagram commutes) if and only if $f_*(\pi_1(X, x_0)) \subset p_*(\pi_1(E, e_0))$.
That said, if $f$ is some finite-sheeted covering map $S^1 \to S^1$ and $p : \Bbb R \to S^1$ the universal covering of $S^1$, then $f_*(\pi_1(S^1))$ would be isomorphic to $\Bbb Z$ (as the only finite index subgroups of $\Bbb Z$ are $n\Bbb Z \cong \Bbb Z$), whereas $p_*(\pi_1(\Bbb R)) \cong 0$. Since $\Bbb Z$ doesn't inject onto the trivial group, there is no such lift.
As the identity map $\text{id} : S^1 \to S^1$ and the two-fold cover $S^1 \to S^1$ are both finite sheeted covering maps, that answer to the question is no.
{"set_name": "stack_exchange", "score": 1, "question_id": 1341268}