The Stacks project

Proof. Suppose given an element $(c_ i) \in \prod _ i A^ p(X_ i \to Y_{a(i)})$. Then given $\beta \in \mathop{\mathrm{CH}}\nolimits _ k(Y)$ we can map this to the element of $\mathop{\mathrm{CH}}\nolimits _{k - p}(X)$ whose restriction to $X_ i$ is $c_ i \cap \beta |_{Y_{a(i)}}$. This works because $\mathop{\mathrm{CH}}\nolimits _{k - p}(X) = \prod _ i \mathop{\mathrm{CH}}\nolimits _{k - p}(X_ i)$. The same construction works after base change by any $Y' \to Y$ locally of finite type and we get $c \in A^ p(X \to Y)$. Thus we obtain a map $\Psi $ from the right hand side of the formula to the left hand side of the formula. Conversely, given $c \in A^ p(X \to Y)$ and an element $\beta _ i \in \mathop{\mathrm{CH}}\nolimits _ k(Y_{a(i)})$ we can consider the element $(c \cap (Y_{a(i)} \to Y)_*\beta _ i)|_{X_ i}$ in $\mathop{\mathrm{CH}}\nolimits _{k - p}(X_ i)$. The same thing works after base change by any $Y' \to Y$ locally of finite type and we get $c_ i \in A^ p(X_ i \to Y_{a(i)})$. Thus we obtain a map $\Phi $ from the left hand side of the formula to the right hand side of the formula. It is immediate that $\Phi \circ \Psi = \text{id}$. For the converse, suppose that $c \in A^ p(X \to Y)$ and $\beta \in \mathop{\mathrm{CH}}\nolimits _ k(Y)$. Say $\Phi (c) = (c_ i)$. Let $j \in J$. Because $c$ commutes with flat pullback we get

Because $c$ commutes with proper pushforward we get

The left hand side is the cycle on $X$ restricting to $(c \cap \beta )|_{X_ i}$ on $X_ i$ for $i \in I$ with $a(i) = j$ and $0$ else. The right hand side is a cycle on $X$ whose restriction to $X_ i$ is $c_ i \cap \beta |_{Y_ j}$ for $i \in I$ with $a(i) = j$. Thus $c \cap \beta = \Psi ((c_ i))$ as desired. $\square$