This proof is the same as the proof of Lemma \ref{lemma-from-functor-to-weil-classical}; we urge the reader to read the proof of that lemma instead. \medskip\noindent We obtain a contravariant functor from the category of smooth projective schemes over $k$ 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 \ref{homology-example-graded-vector-spaces} we conclude that $H^*(X)$ is a graded commutative algebra! Hence we get our datum (D1). \medskip\noindent 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 $-2$. Our datum (D0) is the vector space $F(1) = G^{-2}(\mathbf{1}(1))$. Since $G$ is a symmetric monoidal functor we see that $F(n) = G^{-2n}(\mathbf{1}(n))$ for all $n \in \mathbf{Z}$. It follows that $$ H^{2r}(X)(r) = G^{2r}(h(X)) \otimes G^{-2r}(\mathbf{1}(r)) = G^0(h(X)(r)) $$ a formula we will frequently use below. \medskip\noindent Let $X$ be a smooth projective scheme over $k$. By Lemma \ref{lemma-composition-correspondences} we have $$ \CH^r(X) \otimes \mathbf{Q} = \text{Corr}^r(\Spec(k), X) = \Hom(\mathbf{1}(-r), h(X)) = \Hom(\mathbf{1}, h(X)(r)) $$ Applying the functor $G$ this maps into $\Hom(G(\mathbf{1}), G(h(X)(r)))$. By taking the image of $1$ in $G^0(\mathbf{1}) = F$ into $G^0(h(X)(r)) = H^{2r}(X)(r)$ we obtain $$ \gamma : \CH^r(X) \otimes \mathbf{Q} \longrightarrow H^{2r}(X)(r) $$ This is the datum (D2). \medskip\noindent Let $X$ be a nonempty smooth projective scheme over $k$ which is equidimensional of dimension $d$. By Lemma \ref{lemma-composition-correspondences} we have $$ \Mor(h(X)(d), \mathbf{1}) = \Mor((X, 1, d), (\Spec(k), 1, 0)) = \text{Corr}^{-d}(X, \Spec(k)) = \CH_d(X) $$ Thus the class of the cycle $[X]$ in $\CH_d(X)$ defines a morphism $h(X)(d) \to \mathbf{1}$. Applying $G$ and taking degree $0$ parts we obtain $$ H^{2d}(X)(d) = G^0(h(X)(d)) \longrightarrow G^0(\mathbf{1}) = F $$ This map $\int_X : H^{2d}(X)(d) \to F$ is the datum (D3). \medskip\noindent Let $X$ be a smooth projective scheme over $k$ which is nonempty and equidimensional of dimension $d$. By Lemma \ref{lemma-dual} we know that $h(X)(d)$ is a left dual to $h(X)$. Hence $G(h(X)(d)) = H^*(X) \otimes_F F(d)[2d]$ is a left dual to $H^*(X)$ in the category of graded $F$-vector spaces. Here $[n]$ is the shift functor on graded vector spaces. By Homology, Lemma \ref{homology-lemma-left-dual-graded-vector-spaces} 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 $H^{2d - i}(X)(d) \otimes_F H^i(X) \to F$. In the proof of Lemma \ref{lemma-dual} we have seen that $\epsilon$ is given by $[\Delta]$ via the identifications $$ \Hom(h(X)(d) \otimes h(X), \mathbf{1}) = \text{Corr}^{-d}(X \times X, \Spec(k)) = \CH_d(X \times X) $$ 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). \medskip\noindent Axiom (B) follows from the assumption that $G$ is compatible with tensor structures and our construction of the cup product above. \medskip\noindent Axiom (C). Our construction of $\gamma$ takes a cycle $\alpha$ on $X$, interprets it a correspondence $a$ from $\Spec(k)$ to $X$ of some degree, and then applies $G$. If $f : Y \to X$ is a morphism of nonempty equidimensional smooth projective schemes over $k$, then $f^!\alpha$ is the pushforward (!) of $\alpha$ by the correspondence $[\Gamma_f]$ from $X$ to $Y$, see Lemma \ref{lemma-functor-and-cycles}. Hence $f^!\alpha$ viewed as a correspondence from $\Spec(k)$ to $Y$ is equal to $a \circ [\Gamma_f]$, see Lemma \ref{lemma-composition-correspondences}. Since $G$ is a functor, we conclude $\gamma$ is compatible with pullbacks, i.e., axiom (C)(a) holds. \medskip\noindent Let $f : Y \to X$ be a morphism of nonempty equidimensional smooth projective schemes over $k$ and let $\beta \in \CH^r(Y)$ 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 \ref{lemma-prep-dual}. Let $b$ be $\beta$ viewed as a correspondence from $\Spec(k)$ to $Y$ of degree $r$. Then $f_*\beta$ viewed as a correspondence from $\Spec(k)$ to $X$ is equal to $a^t \circ b$, see Lemmas \ref{lemma-functor-and-cycles} and \ref{lemma-composition-correspondences}. 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 \ref{lemma-prep-dual}. Thus we have axiom (C)(b). \medskip\noindent To prove axiom (C)(c) we use the discussion in Remark \ref{remark-replace-cup-product-classical}. Hence it suffices to prove that $\gamma$ is compatible with exterior products. Let $X$, $Y$ be nonempty smooth projective schemes over $k$ and let $\alpha$, $\beta$ be cycles on them. Denote $a$, $b$ the corresponding correspondences from $\Spec(k)$ to $X$, $Y$. Then $\alpha \times \beta$ corresponds to the correspondence $a \otimes b$ from $\Spec(k)$ 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. \medskip\noindent Axiom (C)(d) follows because the cycle $[\Spec(k)]$ corresponds to the identity morphism on $h(\Spec(k))$. This finishes the proof of the lemma.