Find the winding number of $f$ with $f(z)=f(e^{2\pi i /m} z)$ I have the following problem:

Let $m\geq 1$ be a natural number and $f:S^1\rightarrow S^1$ a
continuous map such that $f(z)=f(e^{2\pi i /m} z)$ $\forall z \in S^1$. Prove that the degree (or winding number) of $f$ is a multiple
of $m$.

I tried to, without explicitly finding an elevation, use the fact that $f(z)=f(e^{2\pi i /m} z)$ without success.
Any help would be appreciated!
 A: Consider the covering projection $p : \mathbb R \to S^1, p(t) = e^{2\pi i t }$.  The winding number $n(f)$ of $f$ is given as follows.

*

*For any closed path $u : [a,b] \to S^1$ (i.e. we have $u(a) = u(b)$) take any lift $l_u : [a,b] \to \mathbb R$. Then $w(u) = l_u(b) - l_u(a)$ is an integer which does not depend on the choice the  lift $l_u$.

*Consider the closed path $f' = f \circ p \mid_{[0,1]} : [0,1] \to S^1$. Then take $n(f) = w(f')$.

We have $f(p(t)) = f(e^{2 \pi i/m}p(t)) = f(e^{2 \pi i/m}e^{2\pi i t}) = f(e^{2\pi i (t + 1/m)})  = f(p(t+1/m))$ for all $t$. That is, $f \circ p$ has period $1/m$. In particular $f'_1 = f \circ p \mid_{[0,1/m]} : [0,1/m] \to S^1$ is a closed path. Lift it to $l_1 : [0,1/m] \to \mathbb R$. Let $d = l_1(1/m) - l_1(0) \in \mathbb Z$. Define
$l : [0,1] \to \mathbb R, l(t)  = l_k(t) = kd + l_1(t -k/m)$ for $t \in [k/m, (k+1)/m]$. This is a well-defined continuous map because for $t = (k+1)/m$ we have
$$l_k((k+1)/m) =  kd + l_1((k+1)/m -k/m) = kd + l_1(1/m) = kd + l_1(1/m) -l_1(0) + l_1(0) \\= (k+1)d + l_1((k+1)/m - (k+1)/m) = l_{k+1}((k+1)/m) .$$
$l$ is a lift of $f'$ because for $t \in [k/m, (k+1)/m]$ we have
$$p(l(t)) = p(kd + l_1(t -k/m)) = p(kd)\cdot p(l_1(t -k/m)) = 1 \cdot f'_1(t -k/m) = f'(t) .$$
Therefore
$$n(f) = w(f') = l(1) - l(0) = l_{m-1}(1) - l_1(0) = (m-1)d + l_1(1- (m-1)/m) - l_1(0) \\ = (m-1)d + l_1(1/m) - l_1(0) = md. $$
