I need to prove that the function $f:\mathbb{C^*} \rightarrow \mathbb{C} ; \exp{(f(z))} = z$ is not continuous using the fundamental group.
I´ve found this Does every continuous map induce a homomorphism on fundamental groups? but im not able to find why the function induces by f is not a group homomorphism.
I have tryed to define the fundamental group of $\mathbb{C^*}$ in $x_0 = -1$ , $\pi_1 (\mathbb{C^*}, -1)$ and using the loop $\alpha : [0,1] \rightarrow \mathbb{C^*} ; t \rightarrow \exp (2 \pi i t)$ and trying to find if any of the group homomorphism properties are not satisfied.
Where the group homomorphism is defined $\pi_1(f,x_o): \pi_1(\mathbb{C^*},x_o) \rightarrow \pi_1(\mathbb{C},f(x_o))$
However I coudn´t find any contradiction, could anyone help me? Thank you.
Note that $\exp$ is a continuous function $\mathbb C \to \mathbb C^*$. Assume that there exists a continuous $f : \mathbb C^* \to \mathbb C$ such that $\exp(f(z)) = z$ for all $z \in \mathbb C^*$. In other words, we have $\exp \circ f = id_{\mathbb C^*}$.
Since $\exp(f(1)) = 1$, we have $f(1) = 2k\pi i$ for some $k \in \mathbb Z$. On the fundamental groups we obtain induced homomorphisms $f_* : \pi_1(\mathbb C^*,1) \to \pi_1(\mathbb C,2k\pi i)$ and $\exp_* : \pi_1(\mathbb C,2k\pi i) \to \pi_1(\mathbb C^*,1)$. We have $\exp_* \circ f_* = (\exp \circ f)_* = id_* = id$. But $\mathbb C$ is contractible, thus $\pi_1(\mathbb C,2k\pi i) = 0$ and therefore $\exp_* \circ f_* = id$ is the zero map on $\pi_1(\mathbb C^*,1)$. This is only possible when $\pi_1(\mathbb C^*,1) = 0$. However, it is well-known that $\pi_1(\mathbb C^*,1) \approx \mathbb Z$. This contradiction shows that no continuous $f$ exists.
Note that there are uncounatbly many functions $f$ such that $\exp(f(z))=z$. None of them is continuous. However, you can find solutions $f$ which are continuous except on some ray starting at $0$ ("branch cuts").