If $f$ is a closed path in $S^1$ at $1$ and if $m \in \Bbb Z$, then $\deg (f^m) = m \deg (f)$
The proof is the answer of the question Question about closed paths in the fundamental group of a circle
I rewrite it and I also have a question on why $\deg(f^m)=m\theta(1)=m\deg(f)$.
Proof.
Consider the following commutative diagram
\begin{array}{ccccccccc} (\mathbb R,\{0,1\}) & \\ \uparrow{\theta} & \searrow{\exp} & \\ (I,\{0,1\}) & \xrightarrow{f} & (S^1,1) & \end{array}
We have $\exp\circ\theta=f$
and $\exp\circ\theta^m=f^m$
and $\deg (f^m)=\deg(e^{2m\pi i\theta})=m\theta(1)=m\deg(f)$
Can anyone explain why $\deg(e^{2m\pi i\theta})=m\theta(1)$?
Thank you.
Writing $\theta : (I,\{0,1\}) \to (\mathbb R,\{0,1\})$ is not correct because it suggests $\theta(\{0,1\}) \subset \{0,1\}$. In fact, $\theta$ is the unique lift of $f$ (lift means $\exp \circ \theta = f$) such that $\theta(0) = 0$. Since $e^{2\pi i \theta(1)} = exp(\theta(1)) = f(1) = 1$, we see that $\theta(1) \in \mathbb Z$. The number $\theta(1)$ is the degree of $f$.
We have $f^m(z) = (f(z))^m$. What is the lift of this map? It is $m\theta$ which is defined as $(m\theta)(t) = m \cdot \theta(t)$. In fact we have $(m\theta)(0) = 0$ and $$exp((m\theta)(t)) = exp(m\theta(t)) = e^{2 \pi i m \theta(t)} = e^{m \cdot 2 \pi i \theta(t)} = (e^{2 \pi i \theta(t)})^m = exp(\theta(t))^m = f^m(t) .$$ This shows $\deg (f^m) = (m\theta)(1) = m \theta(1) = m\deg(f)$.