Processing math: 100%

The Stacks project

Lemma 42.52.2. In Situation 42.50.1 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),

respectively

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 42.49.5. More precisely, let

b : W \to \mathbf{P}^1_ X \quad \text{and}\quad Q \quad \text{and}\quad T' \subset T \subset W_\infty

be as in the proof of Lemma 42.50.2. By definition c_ p(Z \to X, E) = c'_ p(Q) as bivariant operations where the right hand side is the bivariant class constructed in Lemma 42.49.1 using W, b, Q, T'. By Lemma 42.49.5 we have P'_ p(Q) \circ c = c \circ P'_ p(Q), resp. c'_ p(Q) \circ c = c \circ c'_ p(Q) in A^*(Y \times _ X Z \to X) and we conclude. \square


Comments (0)


Your email address will not be published.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.