Is there an algebraic meaning to the relative cohomology of an algebraic curve? Let $X$ be a smooth projective algebraic curve over $\mathbb{C}$. Let $P_1,\ldots,P_n$ be a collection of points on $X$. Then we can speak of its topological relative cohomology $H^1(X,\{P_1,\ldots,P_n\},\mathbb{C})$. Is this naturally isomorphic to some coherent (or even etale) cohomology group?
 A: Updated: I recently learned in Iversen's book Cohomology of sheaves that relative (singular/de Rham) cohomology has a natural sheaf cohomology interpretation: In general for a smooth complex algebraic variety $X$ and a closed smooth subvariety $Y$, let $i:Y\to X$, $j:X\setminus Y\to X$ be inclusions, there is an isomorphism 
$$H^p(X,Y,\mathbb C)\cong H^p(X,j_!j^*\underline{\mathbb C}_X),\tag{1}\label{1}$$ where $\underline{\mathbb C}_X$ is the constant sheaf on $X$. 
This is because there is an exact sequence of sheaves (Hartshorne, exercise II.1.19)
$$0\to j_!j^*\underline{\mathbb C}_X\to \underline{\mathbb C}_X\to i_*i^*\underline{\mathbb C}_X\to 0,$$
which associates to the long exact sequence of cohomology
$$H^{p-1}(X,\mathbb C)\to H^{p-1}(Y,\mathbb C)\to H^p(X,j_!j^*\underline{\mathbb C}_X)\to H^{p}(X,\mathbb C)\to H^p(X,\mathbb C).$$
Compare to the cohomology sequence of the pair $(X,Y)$, we get the desired isomorphism $(\ref{1})$.
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Original Anwser: Let $X$ be a smooth proper curve and $Y=\{p_1,...,p_n\}$.
Using Lefschetz duality, there is an isomorphism
$$H^1(X,Y,\mathbb C)\cong H^{1}(X\setminus Y,\mathbb C)^*.$$
The (singular) cohomology $H^1(X\setminus Y,\mathbb C)$ is a hypercohomology on $X$ (see Voisin's Hodge Theory, vol. I, Cor. 8.19):
$$H^{1}(X\setminus Y,\mathbb C)\cong \mathbb H^{1}(X,\Omega_X^*(\log Y)).$$
Here 
  $$\Omega^*_X(\log Y):=\{0\to \mathcal{O}_X\xrightarrow{\partial} \Omega_X^1(\log Y)\to 0\}$$ 
is a complex of sheaves, and the sheaf $\Omega_X^1(\log Y)$ is the subsheaf of $j_*\Omega_{X\setminus Y}^1$ generated as $\mathcal{O}_X$-module by the meromorphic forms $\frac{dz_i}{z_i}$ for a local coordinate $z_i$ around $p_i$. 
