hcju/stacksqa huggingface dataset:
Recall that $n_{x'}$ is the common value of $\delta(\xi)$ where $\xi$ is the generic point of an irreducible component passing through $x'$. Moreover, the functions $x' \mapsto n_{x'}$, $x \mapsto n_x$, $y' \mapsto n_{y'}$, and $y \mapsto n_y$ are locally constant. Let $X'_n$, $X_n$, $Y'_n$, and $Y_n$ be the open and closed subscheme of $X'$, $X$, $Y'$, and $Y$ where the function has value $n$. Recall that $[X'] = \sum [X'_n]_n$ and $[Y'] = \sum [Y'_n]_n$. Having said this, it is clear that to prove the lemma we may replace $X'$ by one of its connected components and $X$, $Y'$, $Y$ by the connected component that it maps into. Then we know that $X'$, $X$, $Y'$, and $Y$ are $\delta$-equidimensional in the sense that each irreducible component has the same $\delta$-dimension. Say $n'$, $n$, $m'$, and $m$ is this common value for $X'$, $X$, $Y'$, and $Y$. The last assumption means that $n' - m' = n - m$. \medskip\noindent Choose a factorization $f = g \circ i$ where $i : X \to P$ is an immersion and $g : P \to Y$ is smooth. As $X$ is connected, we see that the relative dimension of $P \to Y$ at points of $i(X)$ is constant. Hence after replacing $P$ by an open neighbourhood of $i(X)$, we may assume that $P \to Y$ has constant relative dimension and $i : X \to P$ is a closed immersion. Denote $g' : Y' \times_Y P \to Y'$ the base change of $g$ and denote $i' : X' \to Y' \times_Y P$ the base change of $i$. It is clear that $g^*[Y] = [P]$ and $(g')^*[Y'] = [Y' \times_Y P]$. Finally, if $\xi' \in X'$ is a generic point, then $\mathcal{O}_{Y' \times_Y P, i'(\xi)}$ is Cohen-Macaulay. Namely, the local ring map $\mathcal{O}_{Y', f'(\xi)} \to \mathcal{O}_{Y' \times_Y P, i'(\xi)}$ is flat with regular fibre (see Algebra, Section \ref{algebra-section-smooth-overview}), a regular local ring is Cohen-Macaulay (Algebra, Lemma \ref{algebra-lemma-regular-ring-CM}), $\mathcal{O}_{Y', f'(\xi)}$ is Cohen-Macaulay by assumption (4) and we get what we want from Algebra, Lemma \ref{algebra-lemma-CM-goes-up}. Thus we reduce to the case discussed in the next paragraph. \medskip\noindent Assume $f$ is a regular closed immersion and $X'$, $X$, $Y'$, and $Y$ are $\delta$-equidimensional of $\delta$-dimensions $n'$, $n$, $m'$, and $m$ and $m' - n' = m - n$. In this case we obtain the result immediately from Lemma \ref{lemma-gysin-easy}.
The fact that $\mathcal{C}$ is finite locally free follows immediately from More on Algebra, Lemma \ref{more-algebra-lemma-base-change-lci-bis}. Choose a factorization $f = g \circ i$ with $g : P \to Y$ smooth and $i$ an immersion. Then we can factor $f' = g' \circ i'$ where $g' : P' \to Y'$ and $i' : X' \to P'$ the base changes. Picture $$ \xymatrix{ X' \ar[r] \ar[d] & P' \ar[r] \ar[d] & Y' \ar[d] \\ X \ar[r] & P \ar[r] & Y } $$ In particular, we see that the gysin map exists for $f'$. By More on Morphisms, Lemmas \ref{more-morphisms-lemma-get-NL} we have $$ \NL_{X/Y} = \left( \mathcal{C}_{X/P} \to i^*\Omega_{P/Y} \right) $$ where $\mathcal{C}_{X/P}$ is the conormal sheaf of the embedding $i$. Similarly for the primed version. We have $(g')^*i^*\Omega_{P/Y} = (i')^*\Omega_{P'/Y'}$ because $\Omega_{P/Y}$ pulls back to $\Omega_{P'/Y'}$ by Morphisms, Lemma \ref{morphisms-lemma-base-change-differentials}. Also, recall that $(g')^*\mathcal{C}_{X/P} \to \mathcal{C}_{X'/P'}$ is surjective, see Morphisms, Lemma \ref{morphisms-lemma-conormal-functorial-flat}. We deduce that the sheaf $\mathcal{C}$ is canonicallly isomorphic to the kernel of the map $(g')^*\mathcal{C}_{X/P} \to \mathcal{C}_{X'/P'}$ of finite locally free modules. Recall that $i^!$ is defined using $\mathcal{N} = \mathcal{C}_{Z/X}^\vee$ and similarly for $(i')^!$. Thus we have $$ res(i^!) = c_{top}(\mathcal{C}^\vee) \circ (i')^! $$ in $A^*(X' \to P')$ by an application of Lemma \ref{lemma-gysin-excess}. Since finally we have $f^! = i^! \circ g^*$, $(f')^! = (i')^! \circ (g')^*$, and $(g')^* = res(g^*)$ we conclude.
Observe that the morphism $b$ is a local complete intersection morphism by More on Algebra, Lemma \ref{more-algebra-lemma-blowup-regular-sequence} and hence the statement makes sense. Since $Z \to X$ is a regular immersion (and hence a fortiori quasi-regular) we see that $\mathcal{C}_{Z/X}$ is finite locally free and the map $\text{Sym}^*(\mathcal{C}_{Z/X}) \to \mathcal{C}_{Z/X, *}$ is an isomorphism, see Divisors, Lemma \ref{divisors-lemma-quasi-regular-immersion}. Since $E = \text{Proj}(\mathcal{C}_{Z/X, *})$ we conclude that $E = \mathbf{P}(\mathcal{C}_{Z/X})$ is a projective space bundle over $Z$. Thus $E \to Z$ is smooth and certainly a local complete intersection morphism. Thus Lemma \ref{lemma-compare-gysin-base-change} applies and we see that $$ res(b^!) = c_{top}(\mathcal{C}^\vee) \circ \pi^! $$ with $\mathcal{C}$ as in the statement there. Of course $\pi^* = \pi^!$ by Lemma \ref{lemma-lci-gysin-flat}. It remains to show that $\mathcal{F}$ is equal to the kernel $\mathcal{C}$ of the map $H^{-1}(j^*\NL_{X'/X}) \to H^{-1}(\NL_{E/Z})$. \medskip\noindent Since $E \to Z$ is smooth we have $H^{-1}(\NL_{E/Z}) = 0$, see More on Morphisms, Lemma \ref{more-morphisms-lemma-NL-smooth}. Hence it suffices to show that $\mathcal{F}$ can be identified with $H^{-1}(j^*\NL_{X'/X})$. By More on Morphisms, Lemmas \ref{more-morphisms-lemma-get-triangle-NL} and \ref{more-morphisms-lemma-NL-immersion} we have an exact sequence $$ 0 \to H^{-1}(j^*\NL_{X'/X}) \to H^{-1}(\NL_{E/X}) \to \mathcal{C}_{E/X'} \to \ldots $$ By the same lemmas applied to $E \to Z \to X$ we obtain an isomorphism $\pi^*\mathcal{C}_{Z/X} = H^{-1}(\pi^*\NL_{Z/X}) \to H^{-1}(\NL_{E/X})$. Thus we conclude.
The first statement follows from More on Morphisms, Lemma \ref{more-morphisms-lemma-morphism-regular-schemes-is-lci}. Observe that $f^![Y] = [X]$, see Lemma \ref{lemma-lci-gysin-easy}. Write $\alpha = ch(\alpha') \cap [Y]$ and $\beta = ch(\beta') \cap [Y]$ where $\alpha', \beta' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ as in Section \ref{section-intersection-regular}. Setting $c = ch(\alpha')$ and $c' = ch(\beta')$ we find $\alpha \cdot \beta = c \cap c' \cap [Y]$ by construction. By Lemma \ref{lemma-lci-gysin-commutes} we know that $f^!$ commutes with both $c$ and $c'$. Hence \begin{align*} f^!(\alpha \cdot \beta) & = f^!(c \cap c' \cap [Y]) \\ & = c \cap c' \cap f^![Y] \\ & = c \cap c' \cap [X] \\ & = (c \cap [X]) \cdot (c' \cap [X]) \\ & = (c \cap f^![Y]) \cdot (c' \cap f^![Y]) \\ & = f^!(\alpha) \cdot f^!(\beta) \end{align*} as desired.
The first statement follows from More on Morphisms, Lemma \ref{more-morphisms-lemma-morphism-regular-schemes-is-lci}. Observe that $f^![Y] = [X]$, see Lemma \ref{lemma-lci-gysin-easy}. Write $\alpha = ch(\alpha') \cap [X]$ and $\beta = ch(\beta') \cap [Y]$ $\alpha' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ and $\beta' \in K_0(\textit{Vect}(Y)) \otimes \mathbf{Q}$ as in Section \ref{section-intersection-regular}. Set $c = ch(\alpha')$ and $c' = ch(\beta')$. We have \begin{align*} f_*(\alpha \cdot f^!\beta) & = f_*(c \cap f^!(c' \cap [Y]_e)) \\ & = f_*(c \cap c' \cap f^![Y]_e) \\ & = f_*(c \cap c' \cap [X]_d) \\ & = f_*(c' \cap c \cap [X]_d) \\ & = c' \cap f_*(c \cap [X]_d) \\ & = \beta \cdot f_*(\alpha) \end{align*} The first equality by the construction of the intersection product. By Lemma \ref{lemma-lci-gysin-commutes} we know that $f^!$ commutes with $c'$. The fact that Chern classes are in the center of the bivariant ring justifies switching the order of capping $[X]$ with $c$ and $c'$. Commuting $c'$ with $f_*$ is allowed as $c'$ is a bivariant class. The final equality is again the construction of the intersection product.
A first remark is that if $\alpha = \sum n_i[X_i]$ and $\beta = \sum m_j[Y_j]$ with $X_i \subset X$ and $Y_j \subset Y$ locally finite families of integral closed subschemes of dimensions $n$ and $m$, then $X_i \times_k Y_j$ is a locally finite collection of closed subschemes of $X \times_k Y$ of dimensions $n + m$ and we can indeed consider $$ \alpha \times \beta = \sum n_i m_j [X_i \times_k Y_j]_{n + m} $$ as a $(n + m)$-cycle on $X \times_k Y$. In this way we obtain an additive map $\times : Z_n(X) \otimes_{\mathbf{Z}} Z_m(Y) \to Z_{n + m}(X \times_k Y)$. The problem is to show that this procedure is compatible with rational equivalence. \medskip\noindent Let $i : X' \to X$ be the inclusion morphism of an integral closed subscheme of dimension $n$. Then flat pullback along the morphism $p' : X' \to \Spec(k)$ is an element $(p')^* \in A^{-n}(X' \to \Spec(k))$ by Lemma \ref{lemma-flat-pullback-bivariant} and hence $c' = i_* \circ (p')^* \in A^{-n}(X \to \Spec(k))$ by Lemma \ref{lemma-push-proper-bivariant}. This produces maps $$ c' \cap - : \CH_m(Y) \longrightarrow \CH_{m + n}(X \times_k Y) $$ which the reader easily sends $[Y']$ to $[X' \times_k Y']_{n + m}$ for any integral closed subscheme $Y' \subset Y$ of dimension $m$. Hence the construction $([X'], [Y']) \mapsto [X' \times_k Y']_{n + m}$ factors through rational equivalence in the second variable, i.e., gives a well defined map $Z_n(X) \otimes_{\mathbf{Z}} \CH_m(Y) \to \CH_{n + m}(X \times_k Y)$. By symmetry the same is true for the other variable and we conclude.
Consider the element $[\Spec(k)] \in \CH_0(\Spec(k))$. We get a map $A^p(X \to \Spec(k)) \to \CH_{-p}(X)$ by sending $c$ to $c \cap [\Spec(k)]$. \medskip\noindent Conversely, suppose we have $\alpha \in \CH_{-p}(X)$. Then we can define $c_\alpha \in A^p(X \to \Spec(k))$ as follows: given $X' \to \Spec(k)$ and $\alpha' \in \CH_n(X')$ we let $$ c_\alpha \cap \alpha' = \alpha \times \alpha' $$ in $\CH_{n - p}(X \times_k X')$. To show that this is a bivariant class we write $\alpha = \sum n_i[X_i]$ as in Definition \ref{definition-cycles}. Consider the composition $$ \coprod X_i \xrightarrow{g} X \to \Spec(k) $$ and denote $f : \coprod X_i \to \Spec(k)$ the composition. Then $g$ is proper and $f$ is flat of relative dimension $-p$. Pullback along $f$ is a bivariant class $f^* \in A^p(\coprod X_i \to \Spec(k))$ by Lemma \ref{lemma-flat-pullback-bivariant}. Denote $\nu \in A^0(\coprod X_i)$ the bivariant class which multiplies a cycle by $n_i$ on the $i$th component. Thus $\nu \circ f^* \in A^p(\coprod X_i \to X)$. Finally, we have a bivariant class $$ g_* \circ \nu \circ f^* $$ by Lemma \ref{lemma-push-proper-bivariant}. The reader easily verifies that $c_\alpha$ is equal to this class and hence is itself a bivariant class. \medskip\noindent To finish the proof we have to show that the two constructions are mutually inverse. Since $c_\alpha \cap [\Spec(k)] = \alpha$ this is clear for one of the two directions. For the other, let $c \in A^p(X \to \Spec(k))$ and set $\alpha = c \cap [\Spec(k)]$. It suffices to prove that $$ c \cap [X'] = c_\alpha \cap [X'] $$ when $X'$ is an integral scheme locally of finite type over $\Spec(k)$, see Lemma \ref{lemma-bivariant-zero}. However, then $p' : X' \to \Spec(k)$ is flat of relative dimension $\dim(X')$ and hence $[X'] = (p')^*[\Spec(k)]$. Thus the fact that the bivariant classes $c$ and $c_\alpha$ agree on $[\Spec(k)]$ implies they agree when capped against $[X']$ and the proof is complete.
By Lemma \ref{lemma-exterior-product-associative} we have $(\alpha \times \beta) \times \gamma = \alpha \times (\beta \times \gamma)$ in $\CH_*(Y \times_k Z \times_k W)$. Consider the closed immersions $$ \Delta_{12} : X \times_k X \longrightarrow X \times_k X \times_k X, \quad (x, x') \mapsto (x, x, x') $$ and $$ \Delta_{23} : X \times_k X \longrightarrow X \times_k X \times_k X, \quad (x, x') \mapsto (x, x', x') $$ Denote $\Delta_{12}^!$ and $\Delta_{23}^!$ the corresponding bivariant classes; observe that $\Delta_{12}^!$ is the restriction (Remark \ref{remark-restriction-bivariant}) of $\Delta^!$ to $X \times_k X \times_k X$ by the map $\text{pr}_{12}$ and that $\Delta_{23}^!$ is the restriction of $\Delta^!$ to $X \times_k X \times_k X$ by the map $\text{pr}_{23}$. Thus clearly the restriction of $\Delta_{12}^!$ by $\Delta_{23}$ is $\Delta^!$ and the restriction of $\Delta_{23}^!$ by $\Delta_{12}$ is $\Delta^!$ too. Thus by Lemma \ref{lemma-gysin-commutes} we have $$ \Delta^! \circ \Delta_{12}^! = \Delta^! \circ \Delta_{23}^!  $$ Now we can prove the lemma by the following sequence of equalities: \begin{align*} (\alpha \cdot \beta) \cdot \gamma & = \Delta^!(\Delta^!(\alpha \times \beta) \times \gamma) \\ & = \Delta^!(\Delta_{12}^!((\alpha \times \beta) \times \gamma)) \\ & = \Delta^!(\Delta_{23}^!((\alpha \times \beta) \times \gamma)) \\ & = \Delta^!(\Delta_{23}^!(\alpha \times (\beta \times \gamma)) \\ & = \Delta^!(\alpha \times \Delta^!(\beta \times \gamma)) \\ & = \alpha \cdot (\beta \cdot \gamma) \end{align*} All equalities are clear from the above except perhaps for the second and penultimate one. The equation $\Delta_{23}^!(\alpha \times (\beta \times \gamma)) = \alpha \times \Delta^!(\beta \times \gamma)$ holds by Remark \ref{remark-commuting-exterior}. Similarly for the second equation.
Denote $g : X \to \Spec(k)$ the structure morphism. The map is the composition of the isomorphisms $$ A^p(X) \to A^{p - d}(X \to \Spec(k)) \to \CH_{d - p}(X) $$ The first is the isomorphism $c \mapsto c \circ g^*$ of Proposition \ref{proposition-compute-bivariant} and the second is the isomorphism $c \mapsto c \cap [\Spec(k)]$ of Lemma \ref{lemma-chow-cohomology-towards-point}. From the proof of Lemma \ref{lemma-chow-cohomology-towards-point} we see that the inverse to the second arrow sends $\alpha \in \CH_{d - p}(X)$ to the bivariant class $c_\alpha$ which sends $\beta \in \CH_*(Y)$ for $Y$ locally of finite type over $k$ to $\alpha \times \beta$ in $\CH_*(X \times_k Y)$. From the proof of Proposition \ref{proposition-compute-bivariant} we see the inverse to the first arrow in turn sends $c_\alpha$ to the bivariant class which sends $\beta \in \CH_*(Y)$ for $Y \to X$ locally of finite type to $\Delta^!(\alpha \times \beta) = \alpha \cdot \beta$. From this the final result of the lemma follows.
Observe that $X \to X \times_k Y$ is an immersion of $X$ into a scheme smooth over $Y$. Hence the gysin map is defined for $f$ (Definition \ref{definition-lci-gysin}). To prove the formula we may decompose $X$ and $Y$ into their connected components, hence we may assume $X$ is smooth over $k$ and equidimensional of dimension $d$ and $Y$ is smooth over $k$ and equidimensional of dimension $e$. Observe that $f^![Y]_e = [X]_d$ (see for example Lemma \ref{lemma-lci-gysin-easy}). Write $\alpha = c \cap [Y]_e$ and $\beta = c' \cap [Y]_e$ and hence $\alpha \cdot \beta = c \cap c' \cap [Y]_e$, see Lemma \ref{lemma-identify-chow-for-smooth}. By Lemma \ref{lemma-lci-gysin-commutes} we know that $f^!$ commutes with both $c$ and $c'$. Hence \begin{align*} f^!(\alpha \cdot \beta) & = f^!(c \cap c' \cap [Y]_e) \\ & = c \cap c' \cap f^![Y]_e \\ & = c \cap c' \cap [X]_d \\ & = (c \cap [X]_d) \cdot (c' \cap [X]_d) \\ & = (c \cap f^![Y]_e) \cdot (c' \cap f^![Y]_e) \\ & = f^!(\alpha) \cdot f^!(\beta) \end{align*} as desired where we have used Lemma \ref{lemma-identify-chow-for-smooth} for $X$ as well. \medskip\noindent An alternative proof can be given by proving that $(f \times f)^!(\alpha \times \beta) = f^!\alpha \times f^!\beta$ and using Lemma \ref{lemma-lci-gysin-composition}.
Observe that $X \to X \times_k Y$ is an immersion of $X$ into a scheme smooth over $Y$. Hence the gysin map is defined for $f$ (Definition \ref{definition-lci-gysin}). To prove the formula we may decompose $X$ and $Y$ into their connected components, hence we may assume $X$ is smooth over $k$ and equidimensional of dimension $d$ and $Y$ is smooth over $k$ and equidimensional of dimension $e$. Observe that $f^![Y]_e = [X]_d$ (see for example Lemma \ref{lemma-lci-gysin-easy}). Write $\alpha = c \cap [X]_d$ and $\beta = c' \cap [Y]_e$, see Lemma \ref{lemma-identify-chow-for-smooth}. We have \begin{align*} f_*(\alpha \cdot f^!\beta) & = f_*(c \cap f^!(c' \cap [Y]_e)) \\ & = f_*(c \cap c' \cap f^![Y]_e) \\ & = f_*(c \cap c' \cap [X]_d) \\ & = f_*(c' \cap c \cap [X]_d) \\ & = c' \cap f_*(c \cap [X]_d) \\ & = \beta \cdot f_*(\alpha) \end{align*} The first equality by the result of Lemma \ref{lemma-identify-chow-for-smooth} for $X$. By Lemma \ref{lemma-lci-gysin-commutes} we know that $f^!$ commutes with $c'$. The commutativity of the intersection product justifies switching the order of capping $[X]_d$ with $c$ and $c'$ (via the lemma). Commuting $c'$ with $f_*$ is allowed as $c'$ is a bivariant class. The final equality is again the lemma.
Recall that $[Y] \cdot [Z] = \Delta^!([Y \times Z])$ where $\Delta^! = c(\Delta : X \to X \times X, \mathcal{T}_{X/k})$ is a higher codimension gysin map (Section \ref{section-gysin-higher-codimension}) with $\mathcal{T}_{X/k} = \SheafHom(\Omega_{X/k}, \mathcal{O}_X)$ locally free of rank $\dim(X)$. We have the equality of schemes $$ Y \cap Z = X \times_{\Delta, (X \times X)} (Y \times Z) $$ and $\dim(Y \times Z) = \dim(Y) + \dim(Z)$ and hence conditions  (1), (2), and (3) of Lemma \ref{lemma-gysin-easy} hold. Finally, if $\xi \in Y \cap Z$, then we have a flat local homomorphism $$ \mathcal{O}_{Y, \xi} \longrightarrow \mathcal{O}_{Y \times Z, \xi} $$ whose ``fibre'' is $\mathcal{O}_{Z, \xi}$. It follows that if both $\mathcal{O}_{Y, \xi}$ and $\mathcal{O}_{Z, \xi}$ are Cohen-Macaulay, then so is $\mathcal{O}_{Y \times Z, \xi}$, see Algebra, Lemma \ref{algebra-lemma-CM-goes-up}. In this way we see that all the hypotheses of Lemma \ref{lemma-gysin-easy} are satisfied and we conclude.
After decomposing $X$ into connected components we may and do assume $X$ is equidimensional of dimension $d$. Write $\alpha = c \cap [X]_n$ with $x \in A^*(X)$, see Lemma \ref{lemma-identify-chow-for-smooth}. Then $$ i_*(i^!(\alpha)) = i_*(i^!(c \cap [X]_n)) = i_*(c \cap i^![X]_n) = i_*(c \cap [Y]_e) = c \cap i_*[Y]_e = \alpha \cdot [Y]_e $$ The first equality by choice of $c$. Then second equality by Lemma \ref{lemma-lci-gysin-commutes}. The third because $i^![X]_d = [Y]_e$ in $\CH_*(Y)$ (Lemma \ref{lemma-lci-gysin-easy}). The fourth because bivariant classes commute with proper pushforward. The last equality by Lemma \ref{lemma-identify-chow-for-smooth}.
Let $\alpha \in \CH^i(X)$ and $\beta \in \CH^j(X)$. Write $\alpha = ch(\alpha') \cap [X]$ and $\beta = ch(\beta') \cap [X]$ $\alpha', \beta' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ as in Section \ref{section-intersection-regular}. Set $c = ch(\alpha')$ and $c' = ch(\beta')$. Then the intersection product in Section \ref{section-intersection-regular} produces $c \cap c' \cap [X]$. This is the same as $\alpha \cdot \beta$ by Lemma \ref{lemma-identify-chow-for-smooth} (or rather the generalization that $A^i(X) \to \CH^i(X)$, $c \mapsto c \cap [X]$ is an isomorphism for any smooth scheme $X$ over $k$).
Consider $n$ and $m$ cycles $\alpha = \sum_{i \in I} n_i[X_i]$ and $\beta = \sum_{j \in J} m_j[Y_j]$ with $X_i \subset X$ and $Y_j \subset Y$ locally finite families of integral closed subschemes of $\delta$-dimensions $n$ and $m$. Let $K \subset I \times J$ be the set of pairs $(i, j) \in I \times J$ such that $X_i$ or $Y_j$ dominates an irreducible component of $S$. Then $\{X_i \times_S Y_j\}_{(i, j) \in K}$ is a locally finite collection of closed subschemes of $X \times_S Y$ of $\delta$-dimension $n + m - 1$. This means we can indeed consider $$ \alpha \times \beta = \sum\nolimits_{(i, j) \in K} n_i m_j [X_i \times_S Y_j]_{n + m - 1} $$ as a $(n + m - 1)$-cycle on $X \times_S Y$. In this way we obtain an additive map $\times : Z_n(X) \otimes_{\mathbf{Z}} Z_m(Y) \to Z_{n + m}(X \times_S Y)$. The problem is to show that this procedure is compatible with rational equivalence. \medskip\noindent Let $i : X' \to X$ be the inclusion morphism of an integral closed subscheme of $\delta$-dimension $n$ which dominates an irreducible component of $S$. Then $p' : X' \to S$ is flat of relative dimension $n - 1$, see More on Algebra, Lemma \ref{more-algebra-lemma-dedekind-torsion-free-flat}. Hence flat pullback along $p'$ is an element $(p')^* \in A^{-n + 1}(X' \to S)$ by Lemma \ref{lemma-flat-pullback-bivariant} and hence $c' = i_* \circ (p')^* \in A^{-n + 1}(X \to S)$ by Lemma \ref{lemma-push-proper-bivariant}. This produces maps $$ c' \cap - : \CH_m(Y) \longrightarrow \CH_{m + n - 1}(X \times_S Y) $$ which sends $[Y']$ to $[X' \times_S Y']_{n + m - 1}$ for any integral closed subscheme $Y' \subset Y$ of $\delta$-dimension $m$. \medskip\noindent Let $i : X' \to X$ be the inclusion morphism of an integral closed subscheme of $\delta$-dimension $n$ such that the composition $X' \to X \to S$  factors through a closed point $s \in S$. Since $s$ is a closed point of the spectrum of a Dedekind domain, we see that $s$ is an effective Cartier divisor on $S$ whose normal bundle is trivial. Denote $c \in A^1(s \to S)$ the gysin homomorphism, see Lemma \ref{lemma-gysin-bivariant}. The morphism $p' : X' \to s$ is flat of relative dimension $n$. Hence flat pullback along $p'$ is an element $(p')^* \in A^{-n}(X' \to S)$ by Lemma \ref{lemma-flat-pullback-bivariant}. Thus $$ c' = i_* \circ (p')^* \circ c \in A^{-n}(X \to S) $$ by Lemma \ref{lemma-push-proper-bivariant}. This produces maps $$ c' \cap - : \CH_m(Y) \longrightarrow \CH_{m + n - 1}(X \times_S Y) $$ which for any integral closed subscheme $Y' \subset Y$ of $\delta$-dimension $m$ sends $[Y']$ to either $[X' \times_S Y']_{n + m - 1}$ if $Y'$ dominates an irreducible component of $S$ or to $0$ if not. \medskip\noindent From the previous two paragraphs we conclude the construction $([X'], [Y']) \mapsto [X' \times_S Y']_{n + m - 1}$ factors through rational equivalence in the second variable, i.e., gives a well defined map $Z_n(X) \otimes_{\mathbf{Z}} \CH_m(Y) \to \CH_{n + m - 1}(X \times_S Y)$. By symmetry the same is true for the other variable and we conclude.
Consider the element $[S]_1 \in \CH_1(S)$. We get a map $A^p(X \to S) \to \CH_{1 - p}(X)$ by sending $c$ to $c \cap [S]_1$. \medskip\noindent Conversely, suppose we have $\alpha \in \CH_{1 - p}(X)$. Then we can define $c_\alpha \in A^p(X \to S)$ as follows: given $X' \to S$ and $\alpha' \in \CH_n(X')$ we let $$ c_\alpha \cap \alpha' = \alpha \times \alpha' $$ in $\CH_{n - p}(X \times_S X')$. To show that this is a bivariant class we write $\alpha = \sum_{i \in I} n_i[X_i]$ as in Definition \ref{definition-cycles}. In particular the morphism $$ g : \coprod\nolimits_{i \in I} X_i \longrightarrow X $$ is proper. Pick $i \in I$. If $X_i$ dominates an irreducible component of $S$, then the structure morphism $p_i : X_i \to S$ is flat and we have $\xi_i = p_i^* \in A^p(X_i \to S)$. On the other hand, if $p_i$ factors as $p'_i : X_i \to s_i$ followed by the inclusion $s_i \to S$ of a closed point, then we have $\xi_i = (p'_i)^* \circ c_i \in A^p(X_i \to S)$ where $c_i \in A^1(s_i \to S)$ is the gysin homomorphism and $(p'_i)^*$ is flat pullback. Observe that $$ A^p(\coprod\nolimits_{i \in I} X_i \to S) = \prod\nolimits_{i \in I} A^p(X_i \to S) $$ Thus we have $$ \xi = \sum n_i \xi_i \in A^p(\coprod\nolimits_{i \in I} X_i \to S) $$ Finally, since $g$ is proper we have a bivariant class $$ g_* \circ \xi \in A^p(X \to S) $$ by Lemma \ref{lemma-push-proper-bivariant}. The reader easily verifies that $c_\alpha$ is equal to this class (please compare with the proof of Lemma \ref{lemma-exterior-product-well-defined-dim-1}) and hence is itself a bivariant class. \medskip\noindent To finish the proof we have to show that the two constructions are mutually inverse. Since $c_\alpha \cap [S]_1 = \alpha$ this is clear for one of the two directions. For the other, let $c \in A^p(X \to S)$ and set $\alpha = c \cap [S]_1$. It suffices to prove that $$ c \cap [X'] = c_\alpha \cap [X'] $$ when $X'$ is an integral scheme locally of finite type over $S$, see Lemma \ref{lemma-bivariant-zero}. However, either $p' : X' \to S$ is flat of relative dimension $\dim_\delta(X') - 1$ and hence $[X'] = (p')^*[S]_1$ or $X' \to S$ factors as $X' \to s \to S$ and hence $[X'] = (p')^*(s \to S)^*[S]_1$. Thus the fact that the bivariant classes $c$ and $c_\alpha$ agree on $[S]_1$ implies they agree when capped against $[X']$ (since bivariant classes commute with flat pullback and gysin maps) and the proof is complete.