Let $f:X\to Y$ be a finite morphism of regular projective curves over a field $k$. Let $D\in \operatorname{Div}(Y)$ any divisor, then I'd like to know what is the relationship between the following $k$-vector spaces:
$$H^i(Y,\mathscr O_Y(D))$$
$$H^i(X ,f^\ast\mathscr O_Y(D))=H^i(X ,\mathscr O_X(f^\ast D))$$
I mean: what happens to the cohomology group related to divisors once that we use the pullback functor?
I wrote this up for myself once, so here is a proof of Mohan's statement. It's more general than you asked for, but I don't think restricting to the curve case makes it any easier!
Injectivity lemma (cf. [Lazarsfeld, Lem. 4.1.14]). Let $f\colon Y \to X$ be a finite dominant morphism of varieties defined over a field $k$, and assume that $X$ is normal. Assume also that $p = \operatorname{char} k$ does not divide $\deg f$. If $\mathscr{E}$ is a locally free sheaf on $X$, then the canonical morphism of $\Gamma(X,\mathcal{O}_X)$-modules $$ H^i(X,\mathscr{E}) \longrightarrow H^i(Y,f^*\mathscr{E}) $$ is a split injection.
Proof. Consider the induced field extension $K(X) \hookrightarrow K(Y)$ between the function fields of $X$ and $Y$, and let $\operatorname{Tr} \colon K(Y) \to K(X)$ be the corresponding trace map. Since $X$ is normal, $\operatorname{Tr}$ induces a morphism $\alpha\colon f_*\mathcal{O}_Y \to \mathcal{O}_X$, and since $\deg f$ is invertible in $k$, we have a factorization $$ \mathcal{O}_X \overset{f^\#}{\longrightarrow} f_*\mathcal{O}_Y \overset{\alpha}{\longrightarrow} \mathcal{O}_X \xrightarrow{\cdot \frac{1}{\deg f}} \mathcal{O}_X $$ of the identity on $\mathcal{O}_X$. After applying $-\otimes_{\mathcal{O}_X}\mathscr{E}$, we get a factorization $$ \mathscr{E} \longrightarrow f_*\mathcal{O}_Y \otimes_{\mathcal{O}_X} \mathscr{E} \longrightarrow \mathscr{E} $$ of the identity on $\mathscr{E}$. Since $f_*\mathcal{O}_Y \otimes_{\mathcal{O}_X} \mathscr{E} \simeq f_*(f^*\mathscr{E})$ by the projection formula, applying $H^i(X,-)$ and using the fact that $f$ is finite, we have a factorization $$ H^i(X,\mathscr{E}) \longrightarrow H^i(Y,f^*\mathscr{E}) \longrightarrow H^i(X,\mathscr{E}) $$ of the identity on $H^i(X,\mathscr{E})$, giving the split injection $H^i(X,\mathscr{E}) \hookrightarrow H^i(Y,f^*\mathscr{E})$ desired. $\blacksquare$