this question was asked on mathunderflow but no one gave a satisfactory answer (perhaps here it will receive more attention?)


Say that one has 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 one could pull back cycles and if one had a projection formula then one could say that $f\_{\\*}$ is surjective. 
In fact, given a cycle $\alpha \in N\_\\*(Y)$ one could consider $f\_\\*f^\*\alpha = f\_\\*f^\*(\alpha \cdot [Y]) = f\_\\*(f^\*\alpha \cdot [X]) = \alpha \cdot f\_\\*[X] = \alpha$, giving a surjectivity of $f\_\\*$.


We can assume $X$ regular and $Y$ Gorenstein (even over C): can one still say that $f\_\\*$ is surjective?


EDIT: if it helps, we can assume f to be an isomorphism in codimension one.



if f is birational, is the pushforward map on the numerical groups surjective?
As it has been mentioned, it is not entirely clear how one should define numerical equivalence on singular varieties. What can definitely be done (and has been done) is to define 
$N^1$ as the numerical group of Cartier divisors and $N\_1$ as the group of 1-cycles modulo numerical equivalence against Cartier divisors. I suppose one could try to do something similar to higher cycles using complete intersection subvarieties, but I have not seen that done and I think there are several potential problems with that.


Anyway, at the same time, this does not seem a numerical issue for me. It seems to me that one could ask the same question for any reasonably defined group of cycles.


In that case it seems to me that the answer still depends on the definition. But not on how one defines numerical equivalence, but on what coefficients one uses. In particular, I think the statement is true using rational coefficients, but fails using integer coefficients.


To see the latter, let $f:X\to Y$ be an arbitrary (so smooth if you want) morphism onto a smooth projective curve *without a section* (For instance, take $X$ to be a curve also). Now any curve in $X$ not contained in a fiber maps onto $Y$ via a finite morphism of degree $>1$. In other words, the class $d[Y]$ is be in the image for some $d>0$, but not $[Y]$. (I guess you could get a freak accident by getting $d[Y]$ for two relatively prime $d$'s but I am pretty sure one can give examples when that does not happen. For instance when $X$ is also a curve.) I'm also sure that based on this idea one can produce an example with both $X$ and $Y$ having higher dimensions.


To see that it works over $\mathbb Q$, observe that the above argument can be repeated. Take an effective irreducible cycle on $Y$, take its preimage and take a relative complete intersection subvariety of the right codimension. That will have the same dimension as the cycle you started with, maps onto that cycle and the only unknown is its degree over its image, but over $\mathbb Q$ it does not matter.


