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I doubt the following claim, but it seems that the proof of Theorem 10.2 (page 301, and one can download the book from libgen.org) in the book "algebraic geometry: an introduction to birational geometry of algebraic varieties" uses it:

Let $V,W$ be smooth projective varieties, and $h: V \to W$ be a dominant morphism. If $Y \subseteq W$ be an effective divisor, then the dimension of global sections $h^0(V, h^*(Y)) = h^0(W,Y)$.

Li Yutong
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  • You're correct: this is false as stated. But it is true if $h$ has connected fibres. Are you sure that hypothesis is not somewhere in the text, or implied by the context? (By the way, who are the authors?) –  Jul 21 '14 at 07:44
  • Oh really?! would you suggest why it is true when fibres are connected? Though I have not seen this condition was hidden somewhere in the proof, I just curious about the result itself. Thank you in advance!! – Li Yutong Jul 21 '14 at 09:15
  • The author is S. Iitaka, and the book can be download through this http://libgen.org/get.php?md5=420D2666009FEB2B36D8355B334387E4 – Li Yutong Jul 21 '14 at 09:16
  • The place where I got confused is in page 301, proof of theorem 10.2. In the context, it is written "$l_{V^#}(mh^*(sY)) = l_{W^#}(msY)$". – Li Yutong Jul 21 '14 at 09:21
  • Dear Li, I wrote an answer to explain my claims. I don't have time to read through Iitaka's book, though. –  Jul 21 '14 at 09:43

1 Answers1

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Let me explain why

  • the statement in the last paragraph of the question is false as written;
  • it is true under the extra assumption that $h$ has connected fibres.

A counterexample for the statement in the question is the following: let $h:C \rightarrow \mathbf P^1$ be a $2:1$ map from an elliptic curve. Let $L=O(2)$. Then $h^0(L)=3$, but $h^*(L)$ has degree 4, so Riemann--Roch says $h^0(h^*(L))=4$.

On the other hand, now suppose $h$ has connected fibres. Then we have $h_* O_V = O_W$. So the projection formula says that for any line bundle $L$ on $W$, we have $h_*(h^*(L))=L \otimes h_* O_V = L$. Since global sections are unchanged by pushforward, we get $H^0(h^*(L))=H^0(h_*(h^*(L))=H^0(L)$.