# 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?

• You always have a map $H^i(D)\to H^i(f^*D)$. If characteristic of $k=0$, this map is injective, – Mohan Feb 18 '18 at 19:12

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. $\blacksquare$
• Thanks Takumi Murayama for this thorough answer. I am wondering: Lazarsfeld states that $Y$ should be normal also (ie makes us pass to the normalization of $Y$ before doing all of this). Is this an error? – hedgehog enthusiast Oct 20 '19 at 20:49
• @hedgehogenthusiast I think that step is not needed, although the general case follows from the case when $Y$ is normal as Lazarsfeld states. For a reference that does not pass to the case when $Y$ is normal, see Lemma 2.2.7 in this draft of a book by de Fernex, Ein, and Mustaţă. – Takumi Murayama Oct 20 '19 at 23:17