Cohomology and inverse image of divisors 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?
 A: 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. $\blacksquare$
