Let $p_n$ : $S^1 \to S^1$ be the covering map $p_n(z) = z^n$, (complex notation), $n \in \mathbb{Z}$, $n$ not $0$.

Question

Prove there is a lift of $p_m$ through $p_n$ if and only if $m=nk$, for some $k \in \mathbb{Z}$.

Clearly we have to use the Theorem that says: there exists a lift of $f$ (that is, a continuous map $g : Z → C$ for which $p ∘ g = f$ and $g(z) = c$) if and only if the induced homomorphisms $π(f) : π_1(Z, z) → π_1(X, f(z))$ and $π(p) : π_1(C, c) → π_1(X, f(z))$ at the level of fundamental groups satisfy

$π_1(f)(π_1(Z,z)) \subset π_1(p)(π_1(C,c))$

So we have to define those applications between $p_m$ and $p_n$, verify that the lift exists and see why the equality $m=nk$ shows.

Answer 1

You are looking for a map $g : S^1 \to S^1$ such that $p_n \circ g = p_m$.

1. If $m = nk$, then $g = p_k$ obviously has this property because $(p_n \circ p_k)(z) = p_n(p_k(z)) = p_n(z^k) = (z^k)^n = z^{nk} = p_{nk}(z) = p_m(z)$.

2. If there exist such $g$, then it must be a covering map. See for example Exercise 1.3.16 in Hatcher. In that case the number of sheets of $p_n$ (which is $\lvert n \rvert $) times the number of sheets of $g$ equals the number of sheets of $p_m$ (which is $\lvert m \rvert$). Thus $\lvert m \rvert = \lvert n \rvert l$ for some $l \in \mathbb N$. Hence $m = nk$ for some $k \in \mathbb Z$.