The Stacks project

Proof. Given $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$ there exists a $\beta \in \mathop{\mathrm{CH}}\nolimits _{k + 1}(W)$ restricting to the flat pullback of $\alpha$ on $b^{-1}(\mathbf{A}^1_ X)$, see Lemma 42.14.2. A second choice of $\beta$ differs from $\beta$ by a cycle supported on $W_\infty$, see Lemma 42.19.3. Since the normal bundle of the effective Cartier divisor $D_\infty \subset \mathbf{P}^1_ X$ of (42.18.1.1) is trivial, the gysin homomorphism $i_\infty ^*$ kills cycle classes supported on $W_\infty$, see Remark 42.29.6. Hence setting $C \cap \alpha = i_\infty ^*\beta$ is well defined.

Since $W_\infty$ and $W_0 = X \times \{ 0\}$ are the pullbacks of the rationally equivalent effective Cartier divisors $D_0, D_\infty$ in $\mathbf{P}^1_ X$, we see that $i_\infty ^*\beta$ and $i_0^*\beta$ map to the same cycle class on $W$; namely, both represent the class $c_1(\mathcal{O}_{\mathbf{P}^1_ X}(1)) \cap \beta$ by Lemma 42.29.4. By our choice of $\beta$ we have $i_0^*\beta = \alpha$ as cycles on $W_0 = X \times \{ 0\}$, see for example Lemma 42.31.1. Thus we see that $i_{\infty , *}(C \cap \alpha ) = i_{0, *}\alpha$ as stated in the lemma.

Observe that the assumptions on $b$ are preserved by any base change by $X' \to X$ locally of finite type. Hence we get an operation $C \cap - : \mathop{\mathrm{CH}}\nolimits _ k(X') \to \mathop{\mathrm{CH}}\nolimits _ k(W'_\infty )$ by the same construction as above. To see that this family of operations defines a bivariant class, we consider the diagram

for $X$ as indicated and the base change of this diagram for any $X' \to X$. We know that flat pullback and $i_\infty ^*$ are bivariant operations, see Lemmas 42.33.2 and 42.33.3. Then a formal argument (involving huge diagrams of schemes and their chow groups) shows that the dotted arrow is a bivariant operation. $\square$

Proof. Immediate from the construction and the fact that a similar statement holds for flat pullback and $i_\infty ^*$. $\square$

Proof. Set $b' = b \circ g : W' \to \mathbf{P}^1_ X$. Denote $i'_\infty : W'_\infty \to W'$ the inclusion morphism. Denote $g_\infty : W'_\infty \to W_\infty$ the restriction of $g$. Given $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$ choose $\beta ' \in \mathop{\mathrm{CH}}\nolimits _{k + 1}(W')$ restricting to the flat pullback of $\alpha$ on $(b')^{-1}\mathbf{A}^1_ X$. Then $\beta = g_*\beta ' \in \mathop{\mathrm{CH}}\nolimits _{k + 1}(W)$ restricts to the flat pullback of $\alpha$ on $b^{-1}\mathbf{A}^1_ X$. Then $i_\infty ^*\beta = g_{\infty , *}(i'_\infty )^*\beta '$ by Lemma 42.29.8. This and the corresponding fact after base change by morphisms $X' \to X$ locally of finite type, corresponds to the assertion made in the lemma. $\square$

Proof. Let $\beta \in \mathop{\mathrm{CH}}\nolimits _{k + 1}(W)$. Denote $i_0 : X = X \times \{ 0\} \to W$ the closed immersion of the fibre over $0$ in $\mathbf{P}^1$. Then $(W_\infty \to X)_* i_\infty ^* \beta = i_0^*\beta$ in $\mathop{\mathrm{CH}}\nolimits _ k(X)$ because $i_{\infty , *}i_\infty ^*\beta$ and $i_{0, *}i_0^*\beta$ represent the same class on $W$ (for example by Lemma 42.29.4) and hence pushforward to the same class on $X$. The restriction of $\beta$ to $b^{-1}(\mathbf{A}^1_ X)$ restricts to the flat pullback of $i_0^*\beta = (W_\infty \to X)_* i_\infty ^* \beta$ because we can check this after pullback by $i_0$, see Lemmas 42.32.2 and 42.32.4. Hence we may use $\beta$ when computing the image of $(W_\infty \to X)_*i_\infty ^*\beta$ under $C$ and we get the desired result. $\square$

Proof. Consider the commutative diagram

with cartesian squares. For an elemnent $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$ choose $\beta \in \mathop{\mathrm{CH}}\nolimits _{k + 1}(W)$ whose restriction to $b^{-1}(\mathbf{A}^1_ X)$ is the flat pullback of $\alpha$. Then $c \cap \beta$ is a class in $\mathop{\mathrm{CH}}\nolimits _*(W_ Y)$ whose restriction to $b_ Y^{-1}(\mathbf{A}^1_ Y)$ is the flat pullback of $c \cap \alpha$. Next, we have

because $c$ is a bivariant class. This exactly says that $C \cap c \cap \alpha = c \cap C \cap \alpha$. The same argument works after any base change by $X' \to X$ locally of finite type. This proves the lemma. $\square$