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Say that I have a morphism of projective algebraic varieties $f: X \to Y$, which is birational. There is a pushforward of cycles morphism $f_*: N_*(X) \to N_*(Y)$.

Now, if I could pull back cycles and if I had a projection formula then I could say that $f_*$ is surjective. In fact, given a cycle $\alpha \in N_*(Y)$ I could consider $$f_*f^*\alpha = f_*f^*(\alpha \cdot [Y]) = f_*(f^*\alpha \cdot [X]) = \alpha \cdot f_*[X] = \alpha ,$$ giving me surjectivity of $f_*$.

In my situation $X$ is regular and $Y$ is Gorenstein (and I am working over $\mathbb{C}$): can I still say that $f_*$ is surjective?

EDIT: if it helps, I'm happy to assume f to be an isomorphism in codimension one.

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What is $N_\ast (X)$? –  Georges Elencwajg Apr 5 '12 at 13:23
    
I've only seen your comment now. By $N_i(X)$ I mean the group of i-dimensional cycles modulo numerical equivalence. –  Jacob Bell Apr 25 '12 at 11:14
    
Ah, thanks, donkey kong. –  Georges Elencwajg Apr 25 '12 at 12:07
    
This question has an accepted answer on Mathoverflow one year ago. –  user18119 Jan 11 '13 at 15:04

1 Answer 1

I think it suffices to prove surjectivitiy on the level of Picard groups. That is, it suffices to show that $f_\ast:\textrm{Pic}(X) \to \textrm{Pic}(Y)$ is surjective. Given a line bundle $L$ on $Y$, the pull-back $f^\ast L$ is a line bundle on $X$. (Here you use that $f$ is ??)

Now, the projection formula shows that $f_\ast f^\ast L \cong L$. (Here you use that $f$ is birational, and thus of degree $1$ on the generic fibre.)

It remains to see why $f^\ast L$ is a line bundle...

Edit: I thought about it a bit. I think it is easy to see that $f^\ast L$ is a line bundle because you can simply compute the stalk of $f^\ast L$ at a point $x \in X$...

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it is a general fact that the pullback of a locally free sheaf of a certain rank is a locally free sheaf of the same rank (it all boils down to tensor products). But why does it suffice to show it for Pic? –  Jacob Bell Oct 5 '11 at 21:12

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