42.35 Lemmas on bivariant classes
In this section we prove some elementary results on bivariant classes. Here is a criterion to see that an operation passes through rational equivalence.
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Lemma 42.35.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a morphism of schemes locally of finite type over $S$. Let $p \in \mathbf{Z}$. Suppose given a rule which assigns to every locally of finite type morphism $Y' \to Y$ and every $k$ a map
\[ c \cap - : Z_ k(Y') \longrightarrow \mathop{\mathrm{CH}}\nolimits _{k - p}(X') \]
where $Y' = X' \times _ X Y$, satisfying condition (3) of Definition 42.33.1 whenever $\mathcal{L}'|_{D'} \cong \mathcal{O}_{D'}$. Then $c \cap -$ factors through rational equivalence.
Proof.
The statement makes sense because given a triple $(\mathcal{L}, s, i : D \to X)$ as in Definition 42.29.1 such that $\mathcal{L}|_ D \cong \mathcal{O}_ D$, then the operation $i^*$ is defined on the level of cycles, see Remark 42.29.6. Let $\alpha \in Z_ k(X')$ be a cycle which is rationally equivalent to zero. We have to show that $c \cap \alpha = 0$. By Lemma 42.21.1 there exists a cycle $\beta \in Z_{k + 1}(X' \times \mathbf{P}^1)$ such that $\alpha = i_0^*\beta - i_\infty ^*\beta $ where $i_0, i_\infty : X' \to X' \times \mathbf{P}^1$ are the closed immersions of $X'$ over $0, \infty $. Since these are examples of effective Cartier divisors with trivial normal bundles, we see that $c \cap i_0^*\beta = j_0^*(c \cap \beta )$ and $c \cap i_\infty ^*\beta = j_\infty ^*(c \cap \beta )$ where $j_0, j_\infty : Y' \to Y' \times \mathbf{P}^1$ are closed immersions as before. Since $j_0^*(c \cap \beta ) \sim _{rat} j_\infty ^*(c \cap \beta )$ (follows from Lemma 42.21.1) we conclude.
$\square$
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Lemma 42.35.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a morphism of schemes locally of finite type over $S$. Let $p \in \mathbf{Z}$. Suppose given a rule which assigns to every locally of finite type morphism $Y' \to Y$ and every $k$ a map
\[ c \cap - : \mathop{\mathrm{CH}}\nolimits _ k(Y') \longrightarrow \mathop{\mathrm{CH}}\nolimits _{k - p}(X') \]
where $Y' = X' \times _ X Y$, satisfying conditions (1), (2) of Definition 42.33.1 and condition (3) whenever $\mathcal{L}'|_{D'} \cong \mathcal{O}_{D'}$. Then $c \cap -$ is a bivariant class.
Proof.
Let $Y' \to Y$ be a morphism of schemes which is locally of finite type. Let $(\mathcal{L}', s', i' : D' \to Y')$ be as in Definition 42.29.1 with pullback $(\mathcal{N}', t', j' : E' \to X')$ to $X'$. We have to show that $c \cap (i')^*\alpha ' = (j')^*(c \cap \alpha ')$ for all $\alpha ' \in \mathop{\mathrm{CH}}\nolimits _ k(Y')$.
Denote $g : Y'' \to Y'$ the smooth morphism of relative dimension $1$ with $i'' : D'' \to Y''$ and $p : D'' \to D'$ constructed in Lemma 42.32.7. (Warning: $D''$ isn't the full inverse image of $D'$.) Denote $f : X'' \to X'$ and $E'' \subset X''$ their base changes by $X' \to Y'$. Picture
\[ \xymatrix{ & X'' \ar[rr] \ar '[d][dd]_ h & & Y'' \ar[dd]^ g \\ E'' \ar[rr] \ar[dd]_ q \ar[ru]^{j''} & & D'' \ar[dd]^ p \ar[ru]^{i''} & \\ & X' \ar '[r][rr] & & Y' \\ E' \ar[rr] \ar[ru]^{j'} & & D' \ar[ru]^{i'} } \]
By the properties given in the lemma we know that $\beta ' = (i')^*\alpha '$ is the unique element of $\mathop{\mathrm{CH}}\nolimits _{k - 1}(D')$ such that $p^*\beta ' = (i'')^*g^*\alpha '$. Similarly, we know that $\gamma ' = (j')^*(c \cap \alpha ')$ is the unique element of $\mathop{\mathrm{CH}}\nolimits _{k - 1 - p}(E')$ such that $q^*\gamma ' = (j'')^*h^*(c \cap \alpha ')$. Since we know that
\[ (j'')^*h^*(c \cap \alpha ') = (j'')^*(c \cap g^*\alpha ') = c \cap (i'')^*g^*\alpha ' \]
by our assumptions on $c$; note that the modified version of (3) assumed in the statement of the lemma applies to $i''$ and its base change $j''$. We similarly know that
\[ q^*(c \cap \beta ') = c \cap p^*\beta ' \]
We conclude that $\gamma ' = c \cap \beta '$ by the uniqueness pointed out above.
$\square$
Here a criterion for when a bivariant class is zero.
Lemma 42.35.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a morphism of schemes locally of finite type over $S$. Let $c \in A^ p(X \to Y)$. For $Y'' \to Y' \to Y$ set $X'' = Y'' \times _ Y X$ and $X' = Y' \times _ Y X$. The following are equivalent
$c$ is zero,
$c \cap [Y'] = 0$ in $\mathop{\mathrm{CH}}\nolimits _*(X')$ for every integral scheme $Y'$ locally of finite type over $Y$, and
for every integral scheme $Y'$ locally of finite type over $Y$, there exists a proper birational morphism $Y'' \to Y'$ such that $c \cap [Y''] = 0$ in $\mathop{\mathrm{CH}}\nolimits _*(X'')$.
Proof.
The implications (1) $\Rightarrow $ (2) $\Rightarrow $ (3) are clear. Assumption (3) implies (2) because $(Y'' \to Y')_*[Y''] = [Y']$ and hence $c \cap [Y'] = (X'' \to X')_*(c \cap [Y''])$ as $c$ is a bivariant class. Assume (2). Let $Y' \to Y$ be locally of finite type. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(Y')$. Write $\alpha = \sum n_ i [Y'_ i]$ with $Y'_ i \subset Y'$ a locally finite collection of integral closed subschemes of $\delta $-dimension $k$. Then we see that $\alpha $ is pushforward of the cycle $\alpha ' = \sum n_ i[Y'_ i]$ on $Y'' = \coprod Y'_ i$ under the proper morphism $Y'' \to Y'$. By the properties of bivariant classes it suffices to prove that $c \cap \alpha ' = 0$ in $\mathop{\mathrm{CH}}\nolimits _{k - p}(X'')$. We have $\mathop{\mathrm{CH}}\nolimits _{k - p}(X'') = \prod \mathop{\mathrm{CH}}\nolimits _{k - p}(X'_ i)$ where $X'_ i = Y'_ i \times _ Y X$. This follows immediately from the definitions. The projection maps $\mathop{\mathrm{CH}}\nolimits _{k - p}(X'') \to \mathop{\mathrm{CH}}\nolimits _{k - p}(X'_ i)$ are given by flat pullback. Since capping with $c$ commutes with flat pullback, we see that it suffices to show that $c \cap [Y'_ i]$ is zero in $\mathop{\mathrm{CH}}\nolimits _{k - p}(X'_ i)$ which is true by assumption.
$\square$
Lemma 42.35.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a morphism of schemes locally of finite type over $S$. Assume we have disjoint union decompositions $X = \coprod _{i \in I} X_ i$ and $Y = \coprod _{j \in J} Y_ j$ by open and closed subschemes and a map $a : I \to J$ of sets such that $f(X_ i) \subset Y_{a(i)}$. Then
\[ A^ p(X \to Y) = \prod \nolimits _{i \in I} A^ p(X_ i \to Y_{a(i)}) \]
Proof.
Suppose given an element $(c_ i) \in \prod _ i A^ p(X_ i \to Y_{a(i)})$. Then given $\beta \in \mathop{\mathrm{CH}}\nolimits _ k(Y)$ we can map this to the element of $\mathop{\mathrm{CH}}\nolimits _{k - p}(X)$ whose restriction to $X_ i$ is $c_ i \cap \beta |_{Y_{a(i)}}$. This works because $\mathop{\mathrm{CH}}\nolimits _{k - p}(X) = \prod _ i \mathop{\mathrm{CH}}\nolimits _{k - p}(X_ i)$. The same construction works after base change by any $Y' \to Y$ locally of finite type and we get $c \in A^ p(X \to Y)$. Thus we obtain a map $\Psi $ from the right hand side of the formula to the left hand side of the formula. Conversely, given $c \in A^ p(X \to Y)$ and an element $\beta _ i \in \mathop{\mathrm{CH}}\nolimits _ k(Y_{a(i)})$ we can consider the element $(c \cap (Y_{a(i)} \to Y)_*\beta _ i)|_{X_ i}$ in $\mathop{\mathrm{CH}}\nolimits _{k - p}(X_ i)$. The same thing works after base change by any $Y' \to Y$ locally of finite type and we get $c_ i \in A^ p(X_ i \to Y_{a(i)})$. Thus we obtain a map $\Phi $ from the left hand side of the formula to the right hand side of the formula. It is immediate that $\Phi \circ \Psi = \text{id}$. For the converse, suppose that $c \in A^ p(X \to Y)$ and $\beta \in \mathop{\mathrm{CH}}\nolimits _ k(Y)$. Say $\Phi (c) = (c_ i)$. Let $j \in J$. Because $c$ commutes with flat pullback we get
\[ (c \cap \beta )|_{\coprod _{a(i) = j} X_ i} = c \cap \beta |_{Y_ j} \]
Because $c$ commutes with proper pushforward we get
\[ (\coprod \nolimits _{a(i) = j} X_ i \to X)_* ((c \cap \beta )|_{\coprod _{a(i) = j} X_ i}) = c \cap (Y_ j \to Y)_*\beta |_{Y_ j} \]
The left hand side is the cycle on $X$ restricting to $(c \cap \beta )|_{X_ i}$ on $X_ i$ for $i \in I$ with $a(i) = j$ and $0$ else. The right hand side is a cycle on $X$ whose restriction to $X_ i$ is $c_ i \cap \beta |_{Y_ j}$ for $i \in I$ with $a(i) = j$. Thus $c \cap \beta = \Psi ((c_ i))$ as desired.
$\square$
Lemma 42.35.6. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a morphism of schemes locally of finite type over $S$. Let $g : Y' \to Y$ be an envelope (Definition 42.22.1) and denote $X' = Y' \times _ Y X$. Let $p \in \mathbf{Z}$ and let $c' \in A^ p(X' \to Y')$. If the two restrictions
\[ res_1(c') = res_2(c') \in A^ p(X' \times _ X X' \to Y' \times _ Y Y') \]
are equal (see proof), then there exists a unique $c \in A^ p(X \to Y)$ whose restriction $res(c) = c'$ in $A^ p(X' \to Y')$.
Proof.
We have a commutative diagram
\[ \xymatrix{ X' \times _ X X' \ar[d]^{f''} \ar@<1ex>[r]^-a \ar@<-1ex>[r]_-b & X' \ar[d]^{f'} \ar[r]_ h & X \ar[d]^ f \\ Y' \times _ Y Y' \ar@<1ex>[r]^-p \ar@<-1ex>[r]_-q & Y' \ar[r]^ g & Y } \]
The element $res_1(c')$ is the restriction (see Remark 42.33.5) of $c'$ for the cartesian square with morphisms $a, f', p, f''$ and the element $res_2(c')$ is the restriction of $c'$ for the cartesian square with morphisms $b, f', q, f''$. Assume $res_1(c') = res_2(c')$ and let $\beta \in \mathop{\mathrm{CH}}\nolimits _ k(Y)$. By Lemma 42.22.4 we can find a $\beta ' \in \mathop{\mathrm{CH}}\nolimits _ k(Y')$ with $g_*\beta ' = \beta $. Then we set
\[ c \cap \beta = h_*(c' \cap \beta ') \]
To see that this is independent of the choice of $\beta '$ it suffices to show that $h_*(c' \cap (p_*\gamma - q_*\gamma ))$ is zero for $\gamma \in \mathop{\mathrm{CH}}\nolimits _ k(Y' \times _ Y Y')$. Since $c'$ is a bivariant class we have
\[ h_*(c' \cap (p_*\gamma - q_*\gamma )) = h_*(a_*(c' \cap \gamma ) - b_*(c' \cap \gamma )) = 0 \]
the last equality since $h_* \circ a_* = h_* \circ b_*$ as $h \circ a = h \circ b$.
Observe that our choice for $c \cap \beta $ is forced by the requirement that $res(c) = c'$ and the compatibility of bivariant classes with proper pushforward.
Of course, in order to define the bivariant class $c$ we need to construct maps $c \cap -: \mathop{\mathrm{CH}}\nolimits _ k(Y_1) \to \mathop{\mathrm{CH}}\nolimits _{k + p}(Y_1 \times _ Y X)$ for any morphism $Y_1 \to Y$ locally of finite type satisfying the conditions listed in Definition 42.33.1. Denote $Y'_1 = Y' \times _ Y Y_1$, $X_1 = X \times _ Y Y_1$. The morphism $Y'_1 \to Y_1$ is an envelope by Lemma 42.22.3. Hence we can use the base changed diagram
\[ \xymatrix{ X'_1 \times _{X_1} X'_1 \ar[d]^{f''_1} \ar@<1ex>[r]^-{a_1} \ar@<-1ex>[r]_-{b_1} & X'_1 \ar[d]^{f'_1} \ar[r]_{h_1} & X_1 \ar[d]^{f_1} \\ Y'_1 \times _{Y_1} Y'_1 \ar@<1ex>[r]^-{p_1} \ar@<-1ex>[r]_-{q_1} & Y'_1 \ar[r]^{g_1} & Y_1 } \]
and the same arguments to get a well defined map $c \cap - : \mathop{\mathrm{CH}}\nolimits _ k(Y_1) \to \mathop{\mathrm{CH}}\nolimits _{k + p}(X_1)$ as before.
Next, we have to check conditions (1), (2), and (3) of Definition 42.33.1 for $c$. For example, suppose that $t : Y_2 \to Y_1$ is a proper morphism of schemes locally of finite type over $Y$. Denote as above the base changes of the first diagram to $Y_1$, resp. $Y_2$, by subscripts ${}_1$, resp. ${}_2$. Denote $t' : Y'_2 \to Y'_1$, $s : X_2 \to X_1$, and $s' : X'_2 \to X'_1$ the base changes of $t$ to $Y'$, $X$, and $X'$. We have to show that
\[ s_*(c \cap \beta _2) = c \cap t_*\beta _2 \]
for $\beta _2 \in \mathop{\mathrm{CH}}\nolimits _ k(Y_2)$. Choose $\beta '_2 \in \mathop{\mathrm{CH}}\nolimits _ k(Y'_2)$ with $g_{2, *}\beta '_2 = \beta _2$. Since $c'$ is a bivariant class and the diagrams
\[ \vcenter { \xymatrix{ X'_2 \ar[d]_{s'} \ar[r]_{h_2} & X_2 \ar[d]^ s \\ X'_1 \ar[r]^{h_1} & X_1 } } \quad \text{and}\quad \vcenter { \xymatrix{ X'_2 \ar[d]_{s'} \ar[r]_{f'_2} & Y'_2 \ar[d]^{t'} \\ X'_2 \ar[r]^{f'_1} & Y'_1 } } \]
are cartesian we have
\[ s_*(c \cap \beta _2) = s_*(h_{2, *}(c' \cap \beta '_2)) = h_{1, *}s'_*(c' \cap \beta '_2) = h_{1, *}(c' \cap (t'_*\beta '_2)) \]
and the final expression computes $c \cap t_*\beta _2$ by construction: $t'_*\beta '_2 \in \mathop{\mathrm{CH}}\nolimits _ k(Y'_1)$ is a class whose image by $g_{1, *}$ is $t_*\beta _2$. This proves condition (1). The other conditions are proved in the same manner and we omit the detailed arguments.
$\square$
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