Let $X$ be a complete one-dimensional curve over $\Bbb C$ (please add any hypotheses in case I forgot them). Then we have the Abel-Jacobi map $$\mathcal{AJ}: \text{Div}_0 \to \Bbb C^g/ \Lambda$$ defined by the following prescription. Let $D$ be a degree zero divisor. Then we find a chain $\sigma_D$ such that $d\sigma_D= D$ and set $$\mathcal {AJ}: D\mapsto \left(\int_{\sigma_D} \omega_i \right)_{i=1}^g \pmod \Lambda \in \Bbb C^g/ \Lambda.$$
I want to fill in the details of the sketch of a proof that I have that $\mathcal{AJ} (\operatorname{div}(f))=0$. So $f$ gives us a map $X\to\Bbb {CP}^1$, but the claim that I don't really get is that we can compute $$\int_{\sigma_{\operatorname{div}(f)}} \omega = \int_0^\infty f_* \omega$$
How can we pushforward the differential $\omega$? Is this the trace in the field extension $K(X):K(\Bbb{CP}^1)$, where $K(\cdot)$ denotes the function field of the curve? Anyways, from the LHS we should get that the map is zero because the only regular 1-form on $\Bbb{CP}^1$ is the zero form.