how to describe $f_*: \pi(S^1,1)\rightarrow \pi(S^1,1)$ in terms of the isomorphism of $\pi(S^1,1) \cong Z$

Question

can someone help me with this question please?

Let $f:S^1 \rightarrow S^1$ be the mapping defined by $f(z)=z^k$ for some integer $k$. Describe $f_*: \pi(S^1,1)\rightarrow \pi(S^1,1)$ in terms of the isomorphism of $\pi(S^1,1) \cong Z$ Thanks in advance.

Answer 1

I don't if I am right, but I will give it a try. This is what I know-

(1) $\displaystyle f_{*} (<\alpha>) = <f \circ \alpha>$ where $\alpha$ is a loop based at $1$ in $S^1$.

(2) The exponential mapping $\pi: \mathbb{R} \rightarrow S^1, x \mapsto e^{2\pi ix}$ and the path $p_n(s) = ns, 0\leq s\leq1, n \in \mathbb{Z}$, joining $0$ to $n$ in $\mathbb{R}$ provide us with an isomomorphism $\phi: \mathbb{Z} \rightarrow \pi(S^1,1), n\mapsto <\pi\circ p_n> $.

(3) Now say we have a loop $\beta$ based at $1$, then we can find $n_o\in \mathbb{Z} $ such that $\pi\circ p_{n_o}=\beta$, so $\beta= e^{ 2\pi i n_o s}$ and then $f\circ \beta$ would be $e^{2\pi i n_o k s}$ which is then $\phi (n_o k)$.

So does it mean in terms of the isomomorphism it just multilies by $k$ or what?