TITLE: how to see the Gysin map explicitly in an easy situation
QUESTION [10 upvotes]: Let $C$ be a smooth projective curve and let $U \subset C$ be an open affine subset, with closed complement $S$ consisting of a finite number of points. I am trying to see explicitly the Gysin map in algebraic de Rham cohomology
$$
H^0_{dR}(S) \longrightarrow H^2_{dR}(C). 
$$ By explicitly, I mean covering $C$ by $U$ and another affine open subset, say $V$, containing $S$ and computing $H^2_{dR}(C)$ a la Cech. In this way, every element can be represented by a global section of $\Omega^1$ on $U \cap V$. On the other hand, elements of $H^0_{dR}(S)$ are functions on $S$. So I am trying to find a way to produce a differential on $U \cap V$ out of a function on $S$. I have the intuition that logarithmic derivative should play a role here but I am stuck. Can anyone help? Thanks!

REPLY [5 votes]: Let me first give an answer over $\mathbb C$ in the analytic topology. Let $V$ be a small neighborhood of the set $S$, and compute cohomology using the cover $C = U \cup V$. Then $U \cap V$ is a union of $\vert S \vert$ punctured disks. The function on $S$ prescribes a set of residues of a holomorphic 1-form on $U \cap V$, and any two holomorphic 1-forms with the same set of residues are cohomologous. So this is your representative of the de Rham cohomology class.
In general you may choose a point $\infty \notin S$ and set $V = C \setminus \{\infty\}$. Part of the Riemann-Roch theorem is that there exists a meromorphic 1-form on $C$ with simple poles only along $S \cup \{\infty\}$, with residues along $S$ prescribed by the given function on $S$; the residue at $\infty$ is necessarily minus the sum of the other residues. This gives a holomorphic 1-form on $U \cap V$ which represents the de Rham cohomology class.