4
$\begingroup$

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.

  • 0
    What is $N_\ast (X)$?2012-04-05
  • 0
    I've only seen your comment now. By $N_i(X)$ I mean the group of i-dimensional cycles modulo numerical equivalence.2012-04-25
  • 0
    Ah, thanks, donkey kong.2012-04-25
  • 0
    This question has an [accepted answer](http://mathoverflow.net/questions/77684/) on Mathoverflow one year ago.2013-01-11

1 Answers 1