3

Let $f:X \to Y$ be a dominant/surjective morphism of projective schemes and $\mathcal{L}$ an invertible sheaf. Is it true that $H^0(\mathcal{L})=H^0(f^*\mathcal{L})$?

The fact I am not totally sure of is whether $H^0(f^*\mathcal{L})=H^0(f_*f^*\mathcal{L})$? If this is true then the statement follows from the projection formula.

user43198
  • 477
  • 2
  • 8

1 Answers1

4

No, it is not true in general. In fact $$ H^0(X,f^*L) = H^0(Y,f_*f^*L) \cong H^0(Y,L \otimes f_*O_X), $$ so as soon as the natural morphism $O_Y \to f_*O_X$ is not an isomorphism (for instance, if $f$ is finite of degree $d \ge 2$) and $L$ is sufficiently ample, this is not equal to $H^0(Y,L)$.

Sasha
  • 20,727