Proof.
We obtain a contravariant functor from the category of smooth projective varieties to the category of graded $F$-vector spaces by setting $H^*(X) = G(h(X))$. By assumption we have a canonical isomorphism
\[ H^*(X \times Y) = G(h(X \times Y)) = G(h(X) \otimes h(Y)) = G(h(X)) \otimes G(h(Y)) = H^*(X) \otimes H^*(Y) \]
compatible with pullbacks. Using pullback along the diagonal $\Delta : X \to X \times X$ we obtain a canonical map
\[ H^*(X) \otimes H^*(X) = H^*(X \times X) \to H^*(X) \]
of graded vector spaces compatible with pullbacks. This defines a functorial graded $F$-algebra structure on $H^*(X)$. Since $\Delta $ commutes with the commutativity constraint $h(X) \otimes h(X) \to h(X) \otimes h(X)$ (switching the factors) and since $G$ is a functor of symmetric monoidal categories (so compatible with commutativity constraints), and by our convention in Homology, Example 12.17.4 we conclude that $H^*(X)$ is a graded commutative algebra! Hence we get our datum (D1).
Since $\mathbf{1}(1)$ is invertible in the category of motives we see that $G(\mathbf{1}(1))$ is invertible in the category of graded $F$-vector spaces. Thus $\sum _ i \dim _ F G^ i(\mathbf{1}(1)) = 1$. By assumption we only get something nonzero in degree $-2$ and we may choose an isomorphism $F[2] \to G(\mathbf{1}(1))$ of graded $F$-vector spaces. Here and below $F[n]$ means the graded $F$-vector space which has $F$ in degree $-n$ and zero elsewhere. Using compatibility with tensor products, we find for all $n \in \mathbf{Z}$ an isomorphism $F[2n] \to G(\mathbf{1}(n))$ compatible with tensor products.
Let $X$ be a smooth projective variety. By Lemma 45.3.1 we have
\[ \mathop{\mathrm{CH}}\nolimits ^ r(X) \otimes \mathbf{Q} = \text{Corr}^ r(\mathop{\mathrm{Spec}}(k), X) = \mathop{\mathrm{Hom}}\nolimits (\mathbf{1}(-r), h(X)) \]
Applying the functor $G$ we obtain
\[ \gamma : \mathop{\mathrm{CH}}\nolimits ^ r(X) \otimes \mathbf{Q} \longrightarrow \mathop{\mathrm{Hom}}\nolimits (G(\mathbf{1}(-r)), H^*(X)) = H^{2r}(X) \]
This is the datum (D2).
Let $X$ be a smooth projective variety of dimension $d$. By Lemma 45.3.1 we have
\[ \mathop{\mathrm{Mor}}\nolimits (h(X)(d), \mathbf{1}) = \mathop{\mathrm{Mor}}\nolimits ((X, 1, d), (\mathop{\mathrm{Spec}}(k), 1, 0)) = \text{Corr}^{-d}(X, \mathop{\mathrm{Spec}}(k)) = \mathop{\mathrm{CH}}\nolimits _ d(X) \]
Thus the class of the cycle $[X]$ in $\mathop{\mathrm{CH}}\nolimits _ d(X)$ defines a morphism $h(X)(d) \to \mathbf{1}$. Applying $G$ we obtain
\[ H^*(X) \otimes F[-2d] = G(h(X)(d)) \longrightarrow G(\mathbf{1}) = F \]
This map is zero except in degree $0$ where we obtain $\int _ X : H^{2d}(X) \to F$. This is the datum (D3).
Let $X$ be a smooth projective variety of dimension $d$. By Lemma 45.4.9 we know that $h(X)(d)$ is a left dual to $h(X)$. Hence $G(h(X)(d)) = H^*(X) \otimes F[-2d]$ is a left dual to $H^*(X)$ in the category of graded $F$-vector spaces. By Homology, Lemma 12.17.5 we find that $\sum _ i \dim _ F H^ i(X) < \infty $ and that $\epsilon : h(X)(d) \otimes h(X) \to \mathbf{1}$ produces nondegenerate pairings $H^{2d - i}(X) \otimes _ F H^ i(X) \to F$. In the proof of Lemma 45.4.9 we have seen that $\epsilon $ is given by $[\Delta ]$ via the identifications
\[ \mathop{\mathrm{Hom}}\nolimits (h(X)(d) \otimes h(X), \mathbf{1}) = \text{Corr}^{-d}(X \times X, \mathop{\mathrm{Spec}}(k)) = \mathop{\mathrm{CH}}\nolimits _ d(X \times X) \]
Thus $\epsilon $ is the composition of $[X] : h(X)(d) \to \mathbf{1}$ and $h(\Delta )(d) : h(X)(d) \otimes h(X) \to h(X)(d)$. It follows that the pairings above are given by cup product followed by $\int _ X$. This proves axiom (A) parts (a) and (b).
Axiom (B) follows from the assumption that $G$ is compatible with tensor structures and our construction of the cup product above.
Axiom (C). Our construction of $\gamma $ takes a cycle $\alpha $ on $X$, interprets it as a correspondence $a$ from $\mathop{\mathrm{Spec}}(k)$ to $X$ of some degree, and then applies $G$. If $f : Y \to X$ is a morphism of smooth projective varieties, then $f^!\alpha $ is the pushforward (!) of $\alpha $ by the correspondence $[\Gamma _ f]$ from $X$ to $Y$, see Lemma 45.3.6. Hence $f^!\alpha $ viewed as a correspondence from $\mathop{\mathrm{Spec}}(k)$ to $Y$ is equal to $a \circ [\Gamma _ f]$, see Lemma 45.3.1. Since $G$ is a functor, we conclude $\gamma $ is compatible with pullbacks, i.e., axiom (C)(a) holds.
Let $f : Y \to X$ be a morphism of smooth projective varieties and let $\beta \in \mathop{\mathrm{CH}}\nolimits ^ r(Y)$ be a cycle on $Y$. We have to show that
\[ \int _ Y \gamma (\beta ) \cup f^*c = \int _ X \gamma (f_*\beta ) \cup c \]
for all $c \in H^*(X)$. Let $a, a^ t, \eta _ X, \eta _ Y, [X], [Y]$ be as in Lemma 45.3.9. Let $b$ be $\beta $ viewed as a correspondence from $\mathop{\mathrm{Spec}}(k)$ to $Y$ of degree $r$. Then $f_*\beta $ viewed as a correspondence from $\mathop{\mathrm{Spec}}(k)$ to $X$ is equal to $a^ t \circ b$, see Lemmas 45.3.6 and 45.3.1. The displayed equality above holds if we can show that
\[ h(X) = \mathbf{1} \otimes h(X) \xrightarrow {b \otimes 1} h(Y)(r) \otimes h(X) \xrightarrow {1 \otimes a} h(Y)(r) \otimes h(Y) \xrightarrow {\eta _ Y} h(Y)(r) \xrightarrow {[Y]} \mathbf{1}(r - e) \]
is equal to
\[ h(X) = \mathbf{1} \otimes h(X) \xrightarrow {a^ t \circ b \otimes 1} h(X)(r + d - e) \otimes h(X) \xrightarrow {\eta _ X} h(X)(r + d - e) \xrightarrow {[X]} \mathbf{1}(r - e) \]
This follows immediately from Lemma 45.3.9. Thus we have axiom (C)(b).
To prove axiom (C)(c) we use the discussion in Remark 45.7.2. Hence it suffices to prove that $\gamma $ is compatible with exterior products. Let $X$, $Y$ be smooth projective varieties and let $\alpha $, $\beta $ be cycles on them. Denote $a$, $b$ the corresponding correspondences from $\mathop{\mathrm{Spec}}(k)$ to $X$, $Y$. Then $\alpha \times \beta $ corresponds to the correspondence $a \otimes b$ from $\mathop{\mathrm{Spec}}(k)$ to $X \otimes Y = X \times Y$. Hence the requirement follows from the fact that $G$ is compatible with the tensor structures on both sides.
Axiom (C)(d) follows because the cycle $[\mathop{\mathrm{Spec}}(k)]$ corresponds to the identity morphism on $h(\mathop{\mathrm{Spec}}(k))$. This finishes the proof of the lemma.
$\square$
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