I need to prove that the function $g:\mathbb{C}^∗ \rightarrow \mathbb{C}^*$ such that $(g(z))^n=z$ is not continuous using the fundamental group.
I tried to use the argument in the question: How to prove that the complex logarithm is not continuous using the Fundamental Group
Hence, I construct the inverse of $g$ that is $p_n : \mathbb{C}^* \rightarrow \mathbb{C}^* ; z \rightarrow z^n$. And now we have that $p_n$ o $g = id_{\mathbb{C}^*}$.
So because of $p_n(g(1))=1$, we have $g(1)=1$, and in the fundamentamental group we obtain induced homormophism: $g_{} : \pi_1(\mathbb{C}^*,1) \rightarrow \pi_1(\mathbb{C}^*,1)$
$P_n{_{}} : \pi_1(\mathbb{C}^*,1) \rightarrow \pi_1(\mathbb{C}^*,1)$
But now we can't use the fact that $\mathbb{C}$ is contractible.
How could I adapt the answer to that question to my particular problem? Thank you so much in advance.
Suppose that $ g\colon \mathbb{C}^* \to \mathbb{C}^* $ is continuous. Then a group homomorphism $ g_*\colon \pi_1(\mathbb{C}^*, 1) \to \pi_1(\mathbb{C}^*, g(1)) $ is induced. Since $ p_n \circ g = \mathrm{id}_{\mathbb{C}^*} $, $ (p_n)_* \circ g_* = \mathrm{id}_{\pi_1(\mathbb{C}^*, 1)} $. On the other hand, $ \pi_1(\mathbb{C}^*, 1) \cong \mathbb{Z} $ with a generator $ [0, 1] \to \mathbb{C}^* $; $ t \mapsto e^{2\pi it} $, and similarly $ \pi_1(\mathbb{C}^*, g(1)) \cong \mathbb{Z} $. Under this isomorphism, $ (p_n)_* $ is identified with the map $ \mathbb{Z} \to \mathbb{Z} $; $ k \mapsto nk $. If $ n \geq 2 $, this is not surjective, and therefore it is impossible that $ (p_n)_* \circ g_* = \mathrm{id}_{\pi_1(\mathbb{C}^*, 1)} $.