I have the following group homomorphism: $\pi_1(i,z_o):\pi_1(U,z_o)\rightarrow \pi_1(\mathbb{C},z_o)$ where $i$ is the inclusion map and I know that it is the null-homomorphism, i.e.
$\pi_1(i,z_o)([\alpha]) = [C_{z_o}] $ for every $[\alpha]$ in $\pi_1(U,z_o)$.
I have to prove that the 1-form $\omega=\frac{dz}{z}$ is exact on $U$.
I think that the idea of the proof uses the fact that if $U$ is simply connected then $\omega$ is exact; however, I don't know how to see that $U$ is simply connected when only knowing that the inclusion is the null-homomorphism.
As you have said as the inclusion is the null-homomorphism, every loop in U is null-homologous in $\mathbb{C}$.
Now as $U$ is connected $w$ is exact on $U \iff w $ is local exact and for every loop $\alpha :\int_{\alpha}w=0$
$f(z)=\frac{1}{z}$ is homorphic in $\mathbb{C}^*$ by Goursat's theorem $w = \frac{dz}{z}$ is local exact on $\mathbb{C}^*$
$w$ local exact on $\mathbb{C} \Rightarrow$ $w$ local exact on $U$
Finally as for every $\alpha$ in U, $\alpha$ is null-homologous $\int_{\alpha}w=0$ and $w$ is exact.