# if f is birational, is the pushforward map on the numerical groups surjective?

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.

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What is the definition of $N_*(X)$ ? –  Damian Rössler Oct 10 '11 at 10:53
It's the group of cycles modulo numerical equivalence. When your variety is smooth one cane use the intersection pairing to define it. If I'm not mistaken there should be a way to define it over more general schemes by using intersections with cartier divisors. (or perhaps over C one could use the homological equivalence and pretend they are equal). Makes sense? –  Yosemite Sam Oct 10 '11 at 13:10
Suppose $X$ and $Y$ are smooth over a field (otherwise, it is not quite clear how to define numerical equivalence). Then the equations in your post are valid if you work with the Chow theory groups ${\rm CH}^\bullet(X)_{\bf Q}$ and ${\rm CH}^\bullet(Y)_{\bf Q}$ instead of the groups $N_*(X)$ and $N_*(Y)$ (see Fulton's book on intersection theory). In particular the morphism $f_*:{\rm CH}^\bullet(X)_{\bf Q}\to {\rm CH}^\bullet(Y)_{\bf Q}$ is surjective. By definition of Chow theory, this implies that $f_*:N_*(X)\to N_*(Y)$ is also surjective. But maybe your issue lies somewhere else ? –  Damian Rössler Oct 10 '11 at 15:21
I think it should be possible to still use the numerical groups in the setting of the question (although I lack the technical knowledeg to define them). But even with the smoothness assumption (but without $f$ flat) I don't see why you would have surjectivity. Perhaps given a curve $C \in Z_1(Y)$ one could take the preimage and push it down again. Is it that simple? –  Yosemite Sam Oct 10 '11 at 20:47
@konb: yes, if $C$ has dimension at least 1, its generic points live in the open subset where $f$ is an isomorphism, so $Z_i(X)\to Z_i(Y)$ is surjective for $i=1,2$. As you work over an algebraic closed base field, $f_∗$ is also surjective on $Z_0$. –  Qing Liu Oct 11 '11 at 7:03

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.

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Yes if $f : X\to Y$ a surjective morphism of algebraic varieties, then $f_*: Z(X)\to Z(Y)$ is surjective when tensored by $\mathbb Q$: take an irreducible cycle $C$ on $Y$ and its generic point $\xi$, take any closed point in $f^{-1}(\xi)$ and its Zariski closure $D$ in $X$. Then $f_*D$ is a multiple of $C$. In general one can't remove the multiplicity (consider integral $Y$ and $X\to Y$ with generic fiber a conic without rational point). –  Qing Liu Oct 10 '11 at 22:14
all right, I'll lay down all the cards. The case I really only care about $N_1$ and a flopping contraction between threefolds (here codimension of exceptional locus is two). But, I would really like avoid tensoring with Q. (I guess I should have said that from the start right?) –  Yosemite Sam Oct 10 '11 at 22:27
S'andor, everything you write is correct. And one can even write down explicit examples where the "index" gcd(d) is not 1 (your advisor's "Trento examples" are quite close to this, and I have another paper with similar families of 1-parameter families of Calabi-Yau hypersurfaces). But these are not examples of birational morphisms, as the OP asked. Are you suggesting there is a way to embed X --> Y as the contraction of the exceptional locus in a larger ambient variety? –  Jason Starr Oct 11 '11 at 0:12
Jason, I didn't mean to suggest that any such map $X\to Y$ may be embedded as the contraction of the exceptional locus, but I think there should be an example such that. My feeling is that if you just take a sufficiently general contraction, there is no reason that the exceptional locus over its image should have a section. You probably know more about this than I actually. Anyway, given the additional requirement of konb given in his comment above that it should be a flopping contraction, I am not sure that it can happen, but I am also not sure that it can't. –  Sándor Kovács Oct 11 '11 at 6:55

The answer is No. Consider f a blowup of Y along a smooth subvariety of codimension $\geq 2$. If E is the exceptional divisor of f then $f_*(E)=0$.

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That's true , but note that the question asks about surjectivity, not injectivity. –  Artie Prendergast-Smith Oct 10 '11 at 19:46