The Stacks project

Lemma 42.51.2. In the situation of Definition 42.49.4 assume $P_ p(Z \to X, E)$, resp. $c_ p(Z \to X, E)$ is defined. Let $Y \to X$ be locally of finite type and $c \in A^*(Y \to X)$. Then

\[ P_ p(Z \to X, E) \circ c = c \circ P_ p(Z \to X, E), \]


\[ c_ p(Z \to X, E) \circ c = c \circ c_ p(Z \to X, E) \]

in $A^*(Y \times _ X Z \to X)$.

Proof. This follows from Lemma 42.48.4. Namely, our assumptions say $E$ is represented to a bounded complex $\mathcal{E}^\bullet $ of finite locally free $\mathcal{O}_ X$-modules. Let

\[ b : W \to \mathbf{P}^1_ X \quad \text{and}\quad \mathcal{Q}^\bullet \]

be the blowing up and complex of $\mathcal{O}_ W$-modules constructed in More on Flatness, Section 38.44. Let $T \subset W_\infty $ be the closed subscheme whose existence is averted in More on Flatness, Lemma 38.44.1. Let $T' \subset T$ be the open and closed subscheme such that $\mathcal{Q}^\bullet |_{T'}$ is zero, resp. isomorphic to a finite locally free sheaf of rank $< p$ placed in degree $0$. By definition

\[ c_ p(Z \to X, E) = c'_ p(\mathcal{Q}^\bullet ) \]

as bivariant operations (and not just on cycles over $X$) where the right hand side is the bivariant class constructed in Lemma 42.48.1 using $W, b, \mathcal{Q}^\bullet , T'$. By Lemma 42.48.4 we have

\[ P'_ p(\mathcal{Q}^\bullet ) \circ c = c \circ P'_ p(\mathcal{Q}^\bullet ) \quad \text{resp.}\quad c'_ p(\mathcal{Q}^\bullet ) \circ c = c \circ c'_ p(\mathcal{Q}^\bullet ) \]

in $A^*(Y \times _ X Z \to X)$ and we conclude. $\square$

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Your email address will not be published.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