Lemma 42.61.3 (0FBX)—The Stacks project

Table of contents
Part 3: Topics in Scheme Theory
Chapter 42: Chow Homology and Chern Classes
Section 42.61: Exterior product
Lemma 42.61.3 (cite)

Lemma 42.61.3. Let $k$ be a field. Let $X$ be a scheme locally of finite type over $k$. Let $c \in A^ p(X \to \mathop{\mathrm{Spec}}(k))$. Let $Y \to Z$ be a morphism of schemes locally of finite type over $k$. Let $c' \in A^ q(Y \to Z)$. Then $c \circ c' = c' \circ c$ in $A^{p + q}(X \times _ k Y \to Z)$.

Proof. In the proof of Lemma 42.61.2 we have seen that $c$ is given by a combination of proper pushforward, multiplying by integers over connected components, and flat pullback. Since $c'$ commutes with each of these operations by definition of bivariant classes, we conclude. Some details omitted. $\square$

Comments (2)

Comment #7547 by Hao Peng on July 31, 2022 at 21:29

Shouldn't $c\circ c^\prime$ lie in $A^{p+q}(X\times_kY\to Z)$ ?

Comment #7671 by Stacks Project on August 15, 2022 at 11:56

Good catch! Thanks. Fixed here.

Comments (2)

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