The Stacks project

Remark 42.35.5. 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 $S$. Let $X = \coprod _{i \in I} X_ i$ and $Y = \coprod _{j \in J} Y_ j$ be the decomposition of $X$ and $Y$ into their connected components (the connected components are open as $X$ and $Y$ are locally Noetherian, see Topology, Lemma 5.9.6 and Properties, Lemma 28.5.5). Let $a(i) \in J$ be the index such that $f(X_ i) \subset Y_{a(i)}$. Then $A^ p(X \to Y) = \prod A^ p(X_ i \to Y_{a(i)})$ by Lemma 42.35.4. In this setting it is convenient to set

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

This “completed” bivariant group is the subset

\[ A^*(X \to Y)^\wedge \quad \subset \quad \prod \nolimits _{p \geq 0} A^ p(X) \]

consisting of elements $c = (c_0, c_1, c_2, \ldots )$ such that for each connected component $X_ i$ the image of $c_ p$ in $A^ p(X_ i \to Y_{a(i)})$ is zero for almost all $p$. If $Y \to Z$ is a second morphism, then the composition $A^*(X \to Y) \times A^*(Y \to Z) \to A^*(X \to Z)$ extends to a composition $A^*(X \to Y)^\wedge \times A^*(Y \to Z)^\wedge \to A^*(X \to Z)^\wedge $ of completions. We sometimes call $A^*(X)^\wedge = A^*(X \to X)^\wedge $ the completed bivariant cohomology ring of $X$.


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Your email address will not be published.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 0FE0. Beware of the difference between the letter 'O' and the digit '0'.