hcju/stacksqa huggingface dataset:
We observe that the statement makes sense because $k'$ is a finite separable algebra over $k$ (Varieties, Lemma \ref{varieties-lemma-proper-geometrically-reduced-global-sections}) and hence $\Spec(k')$ is smooth and projective over $k$. The compatibility of $H^*$ with direct sums (Lemmas \ref{lemma-weil-additive} and \ref{lemma-trace-disjoint-union}) shows that it suffices to prove the lemma when $X$ is connected. Hence we may assume $X$ is irreducible and we have to show the equivalence of (1), (2), and (3). Set $d = \dim(X)$. This implies that $k'$ is a field finite separable over $k$ and that $X$ is geometrically irreducible over $k'$, see Varieties, Lemmas \ref{varieties-lemma-proper-geometrically-reduced-global-sections} and \ref{varieties-lemma-baby-stein}. \medskip\noindent By Lemma \ref{lemma-generated-by-separable} we see that the closed points in (3) may be assumed to have separable residue fields over $k$. By axioms (A)(a) and (A)(b) we see that conditions (1) and (3) are equivalent. \medskip\noindent If (2) holds, then pick any closed point $x \in X$ whose residue field is finite separable over $k'$. Then $H^0(\Spec(k')) = H^0(X) \to H^0(x)$ is injective for example by Lemma \ref{lemma-injective}. \medskip\noindent Assume the equivalent conditions (1) and (3) hold. Choose $x_1, \ldots, x_r \in X$ as in (1). Choose a finite separable extension $k''/k'$. By Lemma \ref{lemma-otimes} we have $$ H^0(X) \otimes_{H^0(\Spec(k'))} H^0(\Spec(k'')) = H^0(X \times_{\Spec(k')} \Spec(k'')) $$ Thus in order to show that $H^0(\Spec(k')) \to H^0(X)$ is an isomorphism we may replace $k'$ by $k''$. Thus we may assume $x_1, \ldots, x_r$ are $k'$-rational points (this replaces each $x_i$ with multiple points, so $r$ is increased in this step). By Lemma \ref{lemma-classes-points} $\gamma(x_1) = \gamma(x_2) = \ldots = \gamma(x_r)$. By axiom (A)(b) all the maps $H^0(X) \to H^0(x_i)$ are the same. This means (2) holds. \medskip\noindent Finally, Lemma \ref{lemma-dim-0} implies $H^0(X)$ is a separable $F$-algebra if (1) holds.
By Lemma \ref{lemma-weil-additive} we may assume $Y$ is irreducible and a fortiori equidimensional. If $i$ is odd, then after replacing $Y$ by $Y \times Y$ we find an example where $Y$ is equidimensional and $i = -2l$ for some $l > 0$. Set $X = Y \times (\mathbf{P}^1_k)^l$. Using axiom (B)(a) we obtain $$ H^0(X) \supset H^0(Y) \oplus H^i(Y) \otimes_F H^2(\mathbf{P}^1_k)^{\otimes_F l} $$ with both summands nonzero. Thus it is clear that $H^0(X)$ cannot be isomorphic to $H^0$ of the spectrum of $\Gamma(X, \mathcal{O}_X) = \Gamma(Y, \mathcal{O}_Y)$ as this falls into the first summand.
By the construction of $c^A_i$ we may assume $\mathcal{E}$ has constant rank $r$. Let $p : P \to X$ and $p' : P' \to X$ be the projective bundle associated to $\mathcal{E}$ and $\mathcal{E} \otimes \mathcal{L}$. Then there is an isomorphism $g : P \to P'$ such that $g^*\mathcal{O}_{P'}(1) = \mathcal{O}_P(1) \otimes p^*\mathcal{L}$. See Constructions, Lemma \ref{constructions-lemma-twisting-and-proj}. Thus $$ g^*c_1^A(\mathcal{O}_{P'}(1)) = c_1^A(\mathcal{O}_P(1)) + p^*c_1^A(\mathcal{L}) $$ The desired equality follows formally from this and the definition of Chern classes using equation (\ref{equation-chern-classes}).
Observe that the map $\prod_{i \geq 0} A^i(X) \to \prod_{i \geq 0} A^i(X)$ multiplying by $2^i$ on $A^i(X)$ is a ring map. Hence, since $\psi^2$ is also a ring map, it suffices to prove the formulas for additive generators of $K_0(\textit{Vect}(X))$. Thus we may assume $\alpha = [\mathcal{E}]$ for some finite locally free $\mathcal{O}_X$-module $\mathcal{E}$. By construction of the Chern classes of $\mathcal{E}$ we immediately reduce to the case where $\mathcal{E}$ has constant rank $r$. In this case, we can choose a projective smooth morphism $p : P \to X$ such that restriction $A^*(X) \to A^*(P)$ is injective and such that $p^*\mathcal{E}$ has a finite filtration whose graded parts are invertible $\mathcal{O}_P$-modules $\mathcal{L}_j$, see Lemma \ref{lemma-splitting-principle}. Then $[p^*\mathcal{E}] = \sum [\mathcal{L}_j]$ and hence $\psi^2([p^*\mathcal{E}]) = \sum [\mathcal{L}_j^{\otimes 2}]$ by definition of $\psi^2$. Setting $x_j  = c^A_1(\mathcal{L}_j)$ we have $$ c^A(\alpha) = \prod (1 + x_j) \quad\text{and}\quad c^A(\psi^2(\alpha)) = \prod (1 + 2 x_j) $$ in $\prod A^i(P)$ and we have $$ ch^A(\alpha) = \sum \exp(x_j) \quad\text{and}\quad ch^A(\psi^2(\alpha)) = \sum \exp(2 x_j) $$ in $\prod A^i(P)$. From these formulas the desired result follows.
Recall that we have an isomorphism $$ K_0(\textit{Vect}(X)) \otimes \mathbf{Q} \longrightarrow \CH^*(X) \otimes \mathbf{Q},\quad \alpha \longmapsto ch(\alpha) \cap [X] $$ see Chow Homology, Lemma \ref{chow-lemma-K-tensor-Q}. It is an isomorphism of rings by Chow Homology, Remark \ref{chow-remark-chern-character-K}. We define $\gamma$ by the formula $\gamma(\alpha) = ch^H(\alpha')$ where $ch^H$ is as in Lemma \ref{lemma-chern-classes} and $\alpha' \in K_0(\textit{Vect}(X))$ is such that $ch(\alpha') \cap [X] = \alpha$ in $\CH^*(X) \otimes \mathbf{Q}$. \medskip\noindent The construction $\alpha \mapsto \gamma(\alpha)$ is compatible with pullbacks because both $ch^H$ and taking Chern classes is compatible with pullbacks, see Lemma \ref{lemma-chern-classes} and Chow Homology, Remark \ref{chow-remark-gysin-chern-classes}. \medskip\noindent We still have to see that $\gamma$ is graded. Let $\psi^2 : K_0(\textit{Vect}(X)) \to K_0(\textit{Vect}(X))$ be the second Adams operator, see Chow Homology, Lemma \ref{chow-lemma-second-adams-operator}. If $\alpha \in \CH^i(X)$ and $\alpha' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ is the unique element with $ch(\alpha') \cap [X] = \alpha$, then we have seen in Chow Homology, Section \ref{chow-section-intersection-regular} that $\psi^2(\alpha') = 2^i \alpha'$. Hence we conclude that $ch^H(\alpha') \in H^{2i}(X)(i)$ by Lemma \ref{lemma-adams-and-chern} as desired.
The scheme $X$ is smooth and projective over $k$ and hence we have $K_0(X) = K_0(\textit{Vect}(X))$. See Derived Categories of Schemes, Lemmas \ref{perfect-lemma-resolution-property-ample} and \ref{perfect-lemma-K-is-old-K}. Let $\alpha \in K_0(\text{Vect}(X))$ be an element whose restriction to $Z$ is $[\mathcal{F}]$. By Chow Homology, Lemma \ref{chow-lemma-minus-adams-operator} there exists an element $\alpha^\vee$ which restricts to $\mathcal{C}_{Z/X}^\vee$. By the blow up formula (Chow Homology, Lemma \ref{chow-lemma-blow-up-formula}) we have $$ b^![Z] = b^!i_*[Z] = j_* res(b^!)([Z]) = j_*(c_{r - 1}(\mathcal{F}^\vee) \cap \pi^*[Z]) = j_*(c_{r - 1}(\mathcal{F}^\vee) \cap [E]) $$ where $\mathcal{F}$ is the kernel of the surjection $\pi^*\mathcal{C}_{Z/X} \to \mathcal{C}_{E/X'}$. Observe that $b^*\alpha^\vee - [\mathcal{O}_{X'}(E)]$ is an element of $K_0(\text{Vect}(X'))$ which restricts to $[\pi^*\mathcal{C}_{Z/X}^\vee] - [\mathcal{C}_{E/X'}^\vee] = [\mathcal{F}^\vee]$ on $E$. Since capping with Chern classes commutes with $j_*$ we conclude that the above is equal to $$ c_{r - 1}(b^*\alpha^\vee - [\mathcal{O}_{X'}(E)]) \cap [E] $$ in the chow group of $X'$. Hence we see that setting $$ \theta = c_{r - 1}(b^*\alpha^\vee - [\mathcal{O}_{X'}(E)]) \cap [X'] $$ we get the first relation $\theta \cdot [E] = b^![Z]$ for example by Chow Homology, Lemma \ref{chow-lemma-identify-chow-for-smooth}. For the second relation observe that $$ j^!\theta = j^!(c_{r - 1}(b^*\alpha^\vee - [\mathcal{O}_{X'}(E)]) \cap [X']) = c_{r - 1}(\mathcal{F}^\vee) \cap j^![X'] = c_{r - 1}(\mathcal{F}^\vee) \cap [E] $$ in the chow groups of $E$. To prove that $\pi_*$ of this is equal to $[Z]$ it suffices to prove that the degree of the codimension $r - 1$ cycle $(-1)^{r - 1}c_{r - 1}(\mathcal{F}) \cap [E]$ on the fibres of $\pi$ is $1$. This is a computation we omit.
Let $b : X' \to X$ be the blowing up. By (A7) it suffices to show that $$ b^*(\gamma([Z]) \cup a) = b^*\gamma([Z]) \cup b^*a = 0 $$ By Lemma \ref{lemma-divide-pullback-good-blowing-up} we have $$ b^*\gamma([Z]) = \gamma(b^*[Z]) = \gamma([E] \cdot \theta) = \gamma([E]) \cup \gamma(\theta) $$ Hence because $b^*a$ restricts to zero on $E$ and since $\gamma([E]) = c^H_1(\mathcal{O}_{X'}(E))$ we get what we want from (A4).
Let $X$ be a nonempty smooth projective scheme over $k$ which is equidimensional of dimension $d$. We will show that the graded $F$-vector space $H^*(X)(d)[2d]$ is a left dual to $H^*(X)$. This will prove what we want by Homology, Lemma \ref{homology-lemma-left-dual-graded-vector-spaces}. We are going to use axiom (A5) which in particular says that $$ H^*(X \times X)(d) = \bigoplus H^i(X) \otimes H^j(X)(d) = \bigoplus H^i(X)(d) \otimes H^j(X) $$ Define a map $$ \eta : F \longrightarrow H^*(X \times X)(d) $$ by multiplying by $\gamma([\Delta]) \in H^{2d}(X \times X)(d)$. On the other hand, define a map $$ \epsilon : H^*(X \times X)(d) \longrightarrow H^*(X)(d) \xrightarrow{\lambda} F $$ by first using pullback $\Delta^*$ by the diagonal morphism $\Delta : X \to X \times X$ and then using the $F$-linear map $\lambda : H^{2d}(X)(d) \to F$ of axiom (A6) precomposed by the projection $H^*(X)(d) \to H^{2d}(X)(d)$. In order to show that $H^*(X)(d)$ is a left dual to $H^*(X)$ we have to show that the composition of the maps $$ \eta \otimes 1 : H^*(X) \longrightarrow H^*(X \times X \times X)(d) $$ and $$ 1 \otimes \epsilon : H^*(X \times X \times X)(d) \longrightarrow H^*(X) $$ is the identity. If $a \in H^*(X)$ then we see that the composition maps $a$ to $$ (1 \otimes \lambda)(\Delta_{23}^*(q_{12}^*\gamma([\Delta]) \cup q_3^*a)) = (1 \otimes \lambda)(\gamma([\Delta]) \cup p_2^*a) $$ where $q_i : X \times X \times X \to X$ and $q_{ij} : X \times X \times X \to X \times X$ are the projections, $\Delta_{23} : X \times X \to X \times X \times X$ is the diagonal, and $p_i : X \times X \to X$ are the projections. The equality holds because $\Delta_{23}^*(q_{12}^*\gamma([\Delta]) = \Delta_{23}^*\gamma([\Delta \times X]) = \gamma([\Delta])$ and because $\Delta_{23}^* q_3^*a = p_2^*a$. Since $\gamma([\Delta]) \cup p_1^*a = \gamma([\Delta]) \cup p_2^*a$ (see below) the above simplifies to $$ (1 \otimes \lambda)(\gamma([\Delta]) \cup p_1^*a) = a $$ by our choice of $\lambda$ as desired. The second condition $(\epsilon \otimes 1) \circ (1 \otimes \eta) = \text{id}$ of Categories, Definition \ref{categories-definition-dual} is proved in exactly the same manner. \medskip\noindent Note that $p_1^*a$ and $\text{pr}_2^*a$ restrict to the same cohomology class on $\Delta \subset X \times X$. Moreover we have $\mathcal{C}_{\Delta/X \times X} = \Omega^1_\Delta$ which is the restriction of $p_1^*\Omega^1_X$. Hence Lemma \ref{lemma-A5-A6-imply} implies $\gamma([\Delta]) \cup p_1^*a = \gamma([\Delta]) \cup p_2^*a$ and the proof is complete.
We have $\gamma([\Spec(k)]) = 1 \in H^*(\Spec(k))$ by construction. Since $$ H^0(\Spec(k)) = F,\quad H^0(\Spec(k) \times \Spec(k)) = H^0(\Spec(k)) \otimes_F H^0(\Spec(k)) $$ the map $\int_{\Spec(k)} = \lambda$ of axiom (A6) must send $1$ to $1$ because we have seen that $\int_{\Spec(k) \times \Spec(k)} = \int_{\Spec(k)} \int_{\Spec(k)}$ in Lemma \ref{lemma-trace-product}.
It suffices to prove this on generators for $\CH_*(P)$. Thus it suffices to prove this for a cycle class of the form $\xi^i \cdot p^*\alpha$ where $0 \leq i \leq r - 1$ and $\alpha \in \CH_*(X)$. Note that $p_*(\xi^i \cdot p^*\alpha) = 0$ if $i < r - 1$ and $p_*(\xi^{r - 1} \cdot p^*\alpha) = \alpha$. On the other hand, we have $\gamma(\xi^i \cdot p^*\alpha) = c^i \cup p^*\gamma(\alpha)$ and by the projection formula (Lemma \ref{lemma-pushforward}) we have $$ p_*\gamma(\xi^i \cdot p^*\alpha) = p_*(c^i) \cup \gamma(\alpha) $$ Thus it suffices to show that $p_*c^i = 0$ for $i < r - 1$ and $p_*c^{r - 1} = 1$. Equivalently, it suffices to prove that $\lambda_P : H^{2d + 2r - 2}(P)(d + r - 1) \to F$ defined by the rules $$ \lambda_P(c^i \cup p^*(a)) = \left\{ \begin{matrix} 0 & \text{if} & i < r - 1 \\ \int_X(a) & \text{if} & i = r - 1 \end{matrix} \right. $$ satisfies the condition of axiom (A5). This follows from the computation of the class of the diagonal of $P$ in Lemma \ref{lemma-diagonal-projective-bundle}.
We observe that $$ \Spec(k') \times \Spec(k') = \coprod\nolimits_{\sigma \in \text{Gal}(k'/k)} (\Spec(\sigma) \times \text{id})^{-1} \Delta $$ as cycles on $\Spec(k') \times \Spec(k')$. Our construction of $\gamma$ always sends $[X]$ to $1$ in $H^0(X)$. Thus $1 \otimes 1 = 1 = \sum (\Spec(\sigma) \times \text{id})^*\gamma([\Delta])$. Denote $\lambda : H^0(\Spec(k')) \to F$ the map from axiom (A6), in other words $(\text{id} \otimes \lambda)(\gamma(\Delta)) = 1$ in $H^0(\Spec(k'))$. We obtain \begin{align*} \lambda(1) 1 & = (\text{id} \otimes \lambda)(1 \otimes 1) \\ & = (\text{id} \otimes \lambda)( \sum (\Spec(\sigma) \times \text{id})^*\gamma([\Delta])) \\ & = \sum (\Spec(\sigma) \times \text{id})^*( (\text{id} \otimes \lambda)(\gamma([\Delta])) \\ & = \sum (\Spec(\sigma) \times \text{id})^*(1) \\ & = [k' : k] \end{align*} Since $\lambda$ is another name for $\int_{\Spec(k')}$ (Remark \ref{remark-trace}) the proof is complete.
If $l = 1$ or $l = n - 1$ then $p$ is a projective bundle and the result follows from Lemma \ref{lemma-ok-for-projective-bundle}. In general there exists a morphism $$ h : Y \to X \times \mathbf{G}(l, n) $$ such that both $h$ and $p \circ h$ are compositions of projective space bundles. Namely, denote $\mathbf{G}(1, 2, \ldots, l; n)$ the partial flag variety. Then the morphism $$ \mathbf{G}(1, 2, \ldots, l; n) \to \mathbf{G}(l, n) $$ is a compostion of projective space bundles and similarly the structure morphism $\mathbf{G}(1, 2, \ldots, l; n) \to \Spec(k)$ is of this form. Thus we may set $Y = X \times \mathbf{G}(1, 2, \ldots, l; n)$. Since every cycle on $X \times \mathbf{G}(l, n)$ is the pushforward of a cycle on $Y$, the result for $Y \to X$ and the result for $Y \to X \times \mathbf{G}(l, n)$ imply the result for $p$.
By Lemma \ref{lemma-enough} it suffices to show that $i_*(1) = \gamma([Z])$ if $i : Z \to X$ is a closed immersion of nonempty smooth projective equidimensional schemes over $k$. Say $Z$ has codimension $r$ in $X$. Let $\mathcal{L}$ be a sufficiently ample invertible module on $X$. Choose $n > 0$ and a surjection $$ \mathcal{O}_Z^{\oplus n} \to \mathcal{C}_{Z/X} \otimes \mathcal{L}|_Z $$ This gives a morphism $g : Z \to \mathbf{G}(n - r, n)$ to the Grassmanian over $k$, see Constructions, Section \ref{constructions-section-grassmannian}. Consider the composition $$ Z \to X \times \mathbf{G}(n - r, n) \to X $$ Pushforward along the second morphism is compatible with classes of cycles by Lemma \ref{lemma-grassmanian}. The conormal sheaf $\mathcal{C}$ of the closed immersion $Z \to X \times \mathbf{G}(n - r, n)$ sits in a short exact sequence $$ 0 \to \mathcal{C}_{Z/X} \to \mathcal{C} \to g^*\Omega_{\mathbf{G}(n - r, n)} \to 0 $$ See More on Morphisms, Lemma \ref{more-morphisms-lemma-two-unramified-morphisms-formally-smooth}. Since $\mathcal{C}_{Z/X} \otimes \mathcal{L}|_Z$ is the pull back of a finite locally free sheaf on $\mathbf{G}(n - r, n)$ we conclude that the class of $\mathcal{C}$ in $K_0(Z)$ is the pullback of a class in $K_0(X \times \mathbf{G}(n - r, n))$. Hence we have the property for $Z \to X \times \mathbf{G}(n - r, n)$ and we conclude.
We may replace $k''$ by its normal closure over $k$ which is Galois over $k$, see Fields, Lemma \ref{fields-lemma-normal-closure-galois}. Then $k''$ is Galois over $k'$ as well, see Fields, Lemma \ref{fields-lemma-galois-goes-up}. We deduce we have an isomorphism $$ k' \otimes_k k'' \longrightarrow \prod\nolimits_{\sigma \in \text{Gal}(k''/k')} k'',\quad \eta \otimes \zeta \longmapsto (\eta \sigma(\zeta))_\sigma $$ This produces an isomorphism $\coprod_\sigma \Spec(k'') \to \Spec(k') \times \Spec(k'')$ which on cohomology produces the isomorphism $$ H^*(\Spec(k')) \otimes_F H^*(\Spec(k'')) \to \prod\nolimits_\sigma H^*(\Spec(k'')),\quad a' \otimes a'' \longmapsto (\pi^*a' \cup \Spec(\sigma)^*a'')_\sigma $$ where $\pi : \Spec(k'') \to \Spec(k')$ is the morphism corresponding to the inclusion of $k'$ in $k''$. We conclude the lemma is true by taking $a'' = 1$.
We may replace $X$ by a connected component of $X$ (some details omitted). Thus we may assume $X$ is connected and hence irreducible. Set $k' = \Gamma(X, \mathcal{O}_X) = \Gamma(X', \mathcal{O}_{X'})$; we omit the proof of the equality. Choose a closed point $x' \in X'$ which isn't contained in the exceptional divisor and whose residue field $k''$ is separable over $k$; this is possible by Varieties, Lemma \ref{varieties-lemma-smooth-separable-closed-points-dense}. Denote $x \in X$ the image (whose residue field is equal to $k''$ as well of course). Consider the diagram $$ \xymatrix{ x' \times X' \ar[r] \ar[d] & X' \times X' \ar[d] \\ x \times X \ar[r] & X \times X } $$ The class of the diagonal $\Delta = \Delta_X$ pulls back to the class of the ``diagonal point'' $\delta_x : x \to x \times X$ and similarly for the class of the diagonal $\Delta'$. On the other hand, the diagonal point $\delta_x$ pulls back to the diagonal point $\delta_{x'}$ by the left vertical arrow. Write $\gamma([\Delta]) = \sum \eta_i$ with $\eta_i \in H^i(X) \otimes H^{2d - i}(X)(d)$ and $\gamma([\Delta']) = \sum \eta'_i$ with $\eta'_i \in H^i(X') \otimes H^{2d - i}(X')(d)$. The arguments above show that $\eta_0$ and $\eta'_0$ map to the same class in $$ H^0(x') \otimes_F H^{2d}(X')(d) $$ We have $H^0(\Spec(k')) = H^0(X) = H^0(X')$ by axiom (A8). By Lemma \ref{lemma-injective-H0} this common value maps injectively into $H^0(x')$. We conclude that $\eta_0$ maps to $\eta'_0$ by the map $$ H^0(X) \otimes_F H^{2d}(X)(d) \longrightarrow H^0(X') \otimes_F H^{2d}(X')(d) $$ This means that $\int_X$ is equal to $\int_{X'}$ composed with the pullback map. This proves the lemma.
Let $i : Z \to X$ be as in Lemma \ref{lemma-enough-better}. Consider the diagram $$ \xymatrix{ E \ar[r]_j \ar[d]_\pi & X' \ar[d]^b \\ Z \ar[r]^i & X } $$ Let $\theta \in \CH^{r - 1}(X')$ be as in Lemma \ref{lemma-divide-pullback-good-blowing-up}. Then $\pi_*j^!\theta = [Z]$ in $\CH_*(Z)$ implies that $\pi_*\gamma(j^!\theta) = 1$ by Lemma \ref{lemma-ok-for-projective-bundle} because $\pi$ is a projective space bundle. Hence we see that $$ i_*(1) = i_*(\pi_*(\gamma(j^!\theta))) = b_*j_*(j^*\gamma(\theta)) = b_*(j_*(1) \cup \gamma(\theta)) $$ We have $j_*(1) = \gamma([E])$ by (A9). Thus this is equal to $$ b_*(\gamma([E]) \cup \gamma(\theta)) = b_*(\gamma([E] \cdot \theta)) = b_*(\gamma(b^*[Z])) = b_*b^*\gamma([Z]) = b_*(1) \cup \gamma([Z]) $$ Since $b_*(1) = 1$ by Lemma \ref{lemma-pushforward-blowup} the proof is complete.
Given $G$ and $F[2] \to G(\mathbf{1}(1))$ by setting $H^*(X) = G(h(X))$ we obtain data (D1), (D2), and (D3) satisfying all of (A), (B), and (C) except for possibly (A)(c) and (A)(d), see Lemma \ref{lemma-from-functor-to-weil-classical} and its proof. Observe that assumptions (1) and (2) imply axioms (A)(c) and (A)(d) in the presence of the known axioms (A)(a) and (A)(b). \medskip\noindent Conversely, given $H^*$ we get a functor $G$ by the construction of Lemma \ref{lemma-from-weil-to-functor-classical}. Let $X = \mathbf{P}^1, c_0, c_2$ be as in Example \ref{example-decompose-P1}. We have constructed an isomorphism $1(-1) \to (X, c_2, 0)$ of motives in Lemma \ref{lemma-inverse-h2}. In the proof of Lemma \ref{lemma-from-weil-to-functor-classical} we have seen that $G(1(-1)) = G(X, c_2, 0) = H^2(\mathbf{P}^1)[-2]$. Hence the isomorphism $\int_{\mathbf{P}^1} : H^2(\mathbf{P}^1) \to F$ of axiom (A)(d) gives an isomorphism $G(1(-1)) \to F[-2]$ which determines an isomorphism $F[2] \to G(\mathbf{1}(1))$. Finally, since $G(h(X)) = H^*(X)$ assumptions (1) and (2) follow from axiom (A).
Given $G$ as in (2) by setting $H^*(X) = G(h(X))$ we obtain data (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C), see Lemma \ref{lemma-from-functor-to-weil} and its proof. \medskip\noindent Conversely, given data (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C) we get a functor $G$ as in (2) by the construction of the proof of Lemma \ref{lemma-from-weil-to-functor}. \medskip\noindent We omit the detailed proof that these constructions are inverse to each other.
We have axioms (A), (B) and (C)(a), (C)(c), and (C)(d) of Section \ref{section-axioms} by Lemmas \ref{lemma-poincare-duality}, \ref{lemma-trace-product}, and \ref{lemma-trace-base}. We have axiom (C)(b) by Lemma \ref{lemma-done}. Finally, the additional condition of Definition \ref{definition-weil-cohomology-theory} holds because it is the same as our axiom (A8).
Let $E \subset X$ be locally constructible. We have to show that $f(E)$ is locally constructible too. We will show that $f(E) \cap V$ is constructible for any affine open $V \subset Y$. Thus we reduce to the case where $Y$ is affine. In this case $X$ is quasi-compact. Hence we can write $X = U_1 \cup \ldots \cup U_n$ with each $U_i$ affine open in $X$. If $E \subset X$ is locally constructible, then each $E \cap U_i$ is constructible, see Properties, Lemma \ref{properties-lemma-locally-constructible}. Hence, since $f(E) = \bigcup f(E \cap U_i)$ and since finite unions of constructible sets are constructible, this reduces us to the case where $X$ is affine. In this case the result is Algebra, Theorem \ref{algebra-theorem-chevalley}.
There is an immediate reduction to the case where $X$ and hence $Y$ are affine. Say $X = \Spec(R)$ and $Y = \Spec(A)$. Then $X' = \Spec(A')$, where $A'$ is the integral closure of $R$ in $A$, see Definitions \ref{definition-integral-closure} and \ref{definition-normalization-X-in-Y}. By Algebra, Theorem \ref{algebra-theorem-main-theorem} for every $y \in Y$ at which $f$ is quasi-finite, there exists an open $U'_y \subset X'$ such that $(f')^{-1}(U'_y) \to U'_y$ is an isomorphism. Set $U' = \bigcup U'_y$ where $y \in Y$ ranges over all points where $f$ is quasi-finite. It remains to show that $f$ is quasi-finite at all points of $(f')^{-1}(U')$. If $y \in (f')^{-1}(U')$ with image $x \in X$, then we see that $Y_x \to X'_x$ is an isomorphism in a neighbourhood of $y$. Hence there is no point of $Y_x$ which specializes to $y$, since this is true for $f'(y)$ in $X'_x$, see Lemma \ref{lemma-integral-fibres}. By Lemma \ref{lemma-quasi-finite-at-point-characterize} part (3) this implies $f$ is quasi-finite at $y$.
Condition (6) is our definition of a closed immersion, see Schemes, Definitions \ref{schemes-definition-closed-immersion-locally-ringed-spaces} and \ref{schemes-definition-immersion}. So (6) $\Leftrightarrow$ (1). We have (1) $\Rightarrow$ (2) by Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme}. Trivially (2) $\Rightarrow$ (3). \medskip\noindent Assume (3). Each of the morphisms $\Spec(R_j/I_j) \to \Spec(R_j)$ is a closed immersion, see Schemes, Example \ref{schemes-example-closed-immersion-affines}. Hence $i^{-1}(U_j) \to U_j$ is a homeomorphism onto its image and $i^\sharp|_{U_j}$ is surjective. Hence $i$ is a homeomorphism onto its image and $i^\sharp$ is surjective since this may be checked locally. We conclude that (3) $\Rightarrow$ (4). \medskip\noindent The implication (4) $\Rightarrow$ (1) is Schemes, Lemma \ref{schemes-lemma-characterize-closed-immersions}. The implication (5) $\Rightarrow$ (6) is trivial. And the implication (6) $\Rightarrow$ (5) follows from Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme}.
The equivalence of (1) and (2) is immediate from Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme}. If (1) holds, then there is a closed subspace $i : Z \to X$ with $\mathcal{I} = \Ker(i^\sharp)$ by Schemes, Definition \ref{schemes-definition-closed-subspace} and Example \ref{schemes-example-closed-subspace}. By Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme} this is a closed immersion of schemes and (3) holds. Conversely, if (3) holds, then (2) holds by Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme} (which applies because a closed immersion of schemes is a fortiori a closed immersion of locally ringed spaces).