Let $f:Y\dashrightarrow X$ be a generically finite (and separable) rational map of smooth projective $k$varieties and $x,y\in U(k)$ be two rational points, where $U\subset X$ is an open subset of $X$ over which $f_{}:f^{1}(U)\rightarrow U$ is a finite morphism.
1Can I consider the $0$cycle $f_^*x$ associated to $f_^{1}(x)$ as a $0$cycle $x_f$ of Y ?
2If the first question has a positive answer, is it true that that $x_fy_f \in \rm{CH}_0(Y)$ has degree $0$ ?

$\begingroup$ For question 1, are you asking whether the group homomorphism $f^{1}:Z_0(U) \to Z_0(f^{1}(U))$ maps the subgroup of $X$rationally trivial cycles (i.e., cycles rationally equivalent to zero in $X$) to the subgroup of $Y$rationally trivial cycles? This is what is needed for the composite homomorphism $Z_0(U)\to \text{CH}_0(Y)$ to factor through the image of $Z_0(U)$ in $\text{CH}_0(X)$. $\endgroup$ – Jason Starr Jan 13 '16 at 12:31

1$\begingroup$ Since you are assuming both $X$ and $Y$ are smooth and projective of relative dimension $0$, the morphism $f$ is a "local complete intersection morphism" of relative dimension $0$. Thus there is a refined Gysin pullback morphism $f^!:\text{CH}_0(X)\to \text{CH}_0(Y)$ defined by FultonMacPherson, cf. Chapter 6 of Fulton's "Intersection Theory". In particular, your questions seem to be answered positively by Proposition 6.6 (a) and (b). $\endgroup$ – Jason Starr Jan 13 '16 at 12:52

$\begingroup$ One small note: if you drop the hypothesis that $f$ is a local complete intersection morphism, this can easily fail. For instance, if $X$ is a cone over a smooth plane cubic curve and $Y$ is the blowing up of the vertex of the cone, then $f^{1}$ does not preserve all rational equivalences. $\endgroup$ – Jason Starr Jan 13 '16 at 12:53

$\begingroup$ Thank you very much for your answer. Still, I have a question: is this working for rational maps? $\endgroup$ – user3001 Jan 13 '16 at 12:56

1$\begingroup$ I missed the word "rational" in your question. This should still be okay. As a consequence of the refined Gysin homomorphisms, FultonMacPherson define intersection products that factor through rational equivalence for a smooth variety, e.g., for $Y\times_k X$. So if $\Gamma_f \subset Y\times_k X$ is the closure of the graph of $f$, then the homomorphism $Z_0(X)\to \text{CH}_0(Y)$ by $a \mapsto \text{pr}_{Y,*}(\Gamma_f \cdot \text{pr}_X^*(a))$ factors through $\text{CH}_0(X)$. $\endgroup$ – Jason Starr Jan 13 '16 at 13:01
I am just posting the comments above as an answer, so that this question does not remain unanswered. First of all, if $f$ is a regular morphism, then this follows from Proposition 6.6(a) and (b) of Fulton's book. This requires weaker conditions than smoothness of $X$ and $Y$.
MR1644323 (99d:14003)
Fulton, William
Intersection theory.
Second edition. Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. 2.
SpringerVerlag, Berlin, 1998.
ISBN: 354062046X; 0387985492
14C17 (1402)
In the general case, since $X$ and $Y$ are smooth, there are intersection products on the Chow groups of $X\times_k Y$, as constructed by FultonMacPherson and described in Fulton's book. In particular, defining $\Gamma_f$ to be the closure in $X\times_k Y$ of the graph of $f$ (on its maximal domain of definition), there is a welldefined homomorphism on groups of cycle classes, $$ \text{CH}_d(X) \to \text{CH}_d(Y), \ \ a \mapsto \text{pr}_{Y,*}(\Gamma_f\cdot \text{pr}_X^*(a)).$$