Compatibility of flat pullback of algebraic cycles and proper birational pushforward Let $X \xrightarrow{f} Y \xrightarrow{g} Z$ be morphisms of varieties with composition $h=g \circ f$ such that 


*

*$f$ is proper and birational,

*$g,h$ are flat (of relative dimension $n$).


Is it true that $f_* \circ h^* = g^*$ as morphisms of Chow groups $A_k(Z) \to A_{k+n}(Y)$?
My own thoughts: If $f$ is an isomorphism over the open set $U \subset Y$ and $V \subset Z$ is a subvariety such that all components $W_i$ of $g^{-1}(V)$ intersect $U$, then I think it should be true that $g^* [V] = f_* h^* [V]$. Indeed $h^{-1}(V) = f^{-1}(g^{-1}(V))$ has some components $\tilde W_i$ mapping birationally to $W_i$ (preserving the generic multiplicity of the scheme-theoretic preimages) and some other components $R_j$ which lose dimension under $f$ and are thus killed by the pushforward $f_*$. Overall we have
$$g^* [V] = \sum_i [W_i] = \sum_i f_* [\tilde W_i] = f_* \left( \sum_i [\tilde W_i]+ \sum_j [R_j]\right) = f_* h^* [V].$$
Is it perhaps possible to move all cycles on $Z$ into such a position as for $V$ (or write them as a linear combination of those)?
 A: Usually, I would have given a hint, but Georges Elencwajg just now told me that I should write more detailed answers instead of terse hints. So, I shall try.
Let $Z'\subset Z$ an irreducible subvariety of dimension $k$. If we show that $g^*([Z'])=f_*h^*([Z'])$, we would be done. Let $i:Z'\to Z$ be the inclusion and base change to get $X'\stackrel{f'}{\to}Y'\stackrel{g'}{\to} Z'$ and let $h'=g'\circ f'$. Also, let $i$ denote by abuse of notation, the inclusions $X'\to X, Y'\to Y$. Note first that, $f',g',h'$ have the corresponding properties of $f,g,h$. Then, we have $g^*([Z'])=i_*g'^*([Z'])$ and similarly, $f_*h^*([Z])=i_*f'_*g'^*([Z'])$. This implies, it suffices to show $g'^*([Z'])= f'_*g'^*([Z'])$. But, $g'^*([Z'])=[Y']$ and since $f':X'\to Y'$ is birational, the same applies to the other map. 
Addendum: I assume that $Y$ is normal.Then, let $T\subset Y$ be the smallest closed subset such that $f:X-f^{-1}(T)\to Y-T$ is an isomorphism. The point here  is since $Y$ is normal, this implies for any $p\in T$, $\dim f^{-1}(p)>0$.
Now, let $g^*([Z'])=\sum a_i[Y_i]$ where $Y_i$ are the irreducible components of $g^{-1}(Z')$. I claim that no $Y_i\subset T$. This follows from the fact that since $\dim h^{-1}(Z')=k+n$, we must have $\dim f^{-1}(Y_i)=k+n=\dim Y_i$, but if $Y_i\subset T$, this dimension must be larger, since all points in $T$ have positive dimensional fibers.
Thus, $f^{-1}(Y_i)=X_i\cup G_i$, where $f(G_i)\subset T$ and $f:X_i\to Y_i$ is birational. Thus $h^*([Z'])=\sum a_i X_i+H$ where $H$ is a cycle supported on the $G_i$s. Now, $f_*(h^*([Z'])=\sum a_if_*([X_i])+f_*(H)$. Since $\dim f(G_i)<n+k$, we see that it is zero in $A_{k+n}(Y)$. Also, $f_*([X_i])=Y_i$.
