Induced subgroup of $\pi_1(S^1)$ by $p_n$

Question

Consider the following covering map $p_n: S^1 \to S^1, z \mapsto z^n$. Why is the subgroup of $\pi_1(S^1)$ induced by $p_n$ isomorphic to $n\mathbb{Z}$?

I know that $\pi_1(S^1) \cong \mathbb{Z}$ but how can we say anything about the induced subgroup $(p_n)_* \pi_1(S^1)?$

Answer 1

The subgroup $n \mathbb{Z} < \mathbb{Z} \cong \pi_1(S^1)$ is generated by $$\underbrace{[\gamma] \cdot \cdots \cdot [\gamma]}_n,$$ so one can show this directly by writing down a representative $\gamma$ of a generator of $\pi_1(S^1)$ (the identity will do) and then writing an explicit homotopy $$\underbrace{[\gamma] \cdot \cdots \cdot [\gamma]}_n \simeq [p_n(\gamma)] = [\gamma^n].$$

Answer 2

A generator of $\pi_1(S^1)$ is the class of the path $$ \gamma:I\rightarrow S^1, \qquad \gamma(t)=\cos(\pi t)+i\sin(\pi t). $$ Now $$ (p_n)_*[\gamma]=[p_n\circ\gamma] $$ and $p_n\circ\gamma=\cos(\pi nt)+i\sin(\pi nt)$. Thus visibly $(p_n)_*[\gamma]=[\gamma]^n$ (multiplicative notation). So under the isomorphism $\pi_1(S^1)\simeq\Bbb Z$ we have $(p_n)_*(\pi_1(S^1))=n\Bbb Z$