The Stacks project

Lemma 42.34.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a morphism of schemes locally of finite type over $S$. Assume we have disjoint union decompositions $X = \coprod _{i \in I} X_ i$ and $Y = \coprod _{j \in J} Y_ j$ by open and closed subschemes and a map $a : I \to J$ of sets such that $f(X_ i) \subset Y_{a(i)}$. Then

\[ A^ p(X \to Y) = \prod \nolimits _{i \in I} A^ p(X_ i \to Y_{a(i)}) \]

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

\[ (c \cap \beta )|_{\coprod _{a(i) = j} X_ i} = c \cap \beta |_{Y_ j} \]

Because $c$ commutes with proper pushforward we get

\[ (\coprod \nolimits _{a(i) = j} X_ i \to X)_* ((c \cap \beta )|_{\coprod _{a(i) = j} X_ i}) = c \cap (Y_ j \to Y)_*\beta |_{Y_ j} \]

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FDZ. Beware of the difference between the letter 'O' and the digit '0'.