The Stacks project

Proof. By Lemma 45.4.8 it suffices to construct a functor $G$ on the category of smooth projective schemes over $k$ with morphisms given by correspondences of degree $0$ such that the image of $G(c_2)$ on $G(\mathbf{P}^1)$ is an invertible graded $F$-vector space. Since every smooth projective scheme is canonically a disjoint union of smooth projective varieties, it suffices to construct $G$ on the category whose objects are smooth projective varieties and whose morphisms are correspondences of degree $0$. (Some details omitted.)

Given a smooth projective variety $X$ we set $G(X) = H^*(X)$.

Given a correspondence $c \in \text{Corr}^0(X, Y)$ between smooth projective varieties we consider the map $G(c) : G(X) = H^*(X) \to G(Y) = H^*(Y)$ given by the rule

It is clear that $G(c)$ is additive in $c$ and hence $\mathbf{Q}$-linear. Compatibility of $\gamma $ with pullbacks, pushforwards, and intersection products given by axioms (C)(a), (C)(b), and (C)(c) shows that we have $G(c' \circ c) = G(c') \circ G(c)$ if $c' \in \text{Corr}^0(Y, Z)$. Namely, for $a \in H^*(X)$ we have

with obvious notation. The first equality follows from the definitions. The second equality holds because $\text{pr}^{23, *}_2 \circ \text{pr}^{12}_{2, *} = \text{pr}^{123}_{23, *} \circ \text{pr}^{123, *}_{12}$ as follows immediately from the description of pushforward along projections given in Lemma 45.7.6. The third equality holds by Lemma 45.7.1 and the fact that $H^*$ is a functor. The fourth equalith holds by axiom (C)(a) and the fact that the gysin map agrees with flat pullback for flat morphisms (Chow Homology, Lemma 42.59.5). The fifth equality uses axiom (C)(c) as well as Lemma 45.7.1 to see that $\text{pr}^{23}_{3, *} \circ \text{pr}^{123}_{23, *} = \text{pr}^{13}_{3, *} \circ \text{pr}^{123}_{13, *}$. The sixth equality uses the projection formula from Lemma 45.7.1 as well as axiom (C)(b) to see that $ \text{pr}^{123}_{13, *} \gamma (\text{pr}^{123, *}_{23}c' \cdot \text{pr}^{123, *}_{12}c) = \gamma (\text{pr}^{123}_{13, *}( \text{pr}^{123, *}_{23}c' \cdot \text{pr}^{123, *}_{12}c))$. Finally, the last equality is the definition.

To finish the proof that $G$ is a functor, we have to show identities are preserved. In other words, if $1 = [\Delta ] \in \text{Corr}^0(X, X)$ is the identity in the category of correspondences (see Lemma 45.3.3 and its proof), then we have to show that $G([\Delta ]) = \text{id}$. This follows from the determination of $\gamma ([\Delta ])$ in Lemma 45.7.7 and Lemma 45.7.6. This finishes the construction of $G$ as a functor on smooth projective varieties and correspondences of degree $0$.

It follows from axioms (A)(c) and (A)(d) that $G(\mathop{\mathrm{Spec}}(k)) = H^*(\mathop{\mathrm{Spec}}(k))$ is canonically isomorphic to $F$ as an $F$-algebra. The Künneth axiom (B) shows our functor is compatible with tensor products. Thus our functor is a functor of symmetric monoidal categories.

We still have to check that the image of $G(c_2)$ on $G(\mathbf{P}^1)$ is an invertible graded $F$-vector space (in particular we don't know yet that $G$ extends to $M_ k$). By axiom (A)(d) the map $\int _{\mathbf{P}^1} : H^2(\mathbf{P}^1) \to F$ is an isomorphism. By axiom (A)(b) we see that $\dim _ F H^0(\mathbf{P}^1) = 1$. By Lemma 45.7.8 and axiom (A)(c) we obtain $2 - \dim _ F H^1(\mathbf{P}^1) = c_1(T_{\mathbf{P}^1}) = 2$. Hence $H^1(\mathbf{P}^1) = 0$. Thus

Recall that $1 = c_0 + c_2$ is a decomposition of the identity into a sum of orthogonal idempotents in $\text{Corr}^0(\mathbf{P}^1, \mathbf{P}^1)$, see Example 45.3.7. We have $c_0 = a \circ b$ where $a \in \text{Corr}^0(\mathop{\mathrm{Spec}}(k), \mathbf{P}^1)$ and $b \in \text{Corr}^0(\mathbf{P}^1, \mathop{\mathrm{Spec}}(k))$ and where $b \circ a = 1$ in $\text{Corr}^0(\mathop{\mathrm{Spec}}(k), \mathop{\mathrm{Spec}}(k))$, see proof of Lemma 45.4.4. Since $F = G(\mathop{\mathrm{Spec}}(k))$, it follows from functoriality that $G(c_0)$ is the projector onto the summand $H^0(\mathbf{P}^1) \subset G(\mathbf{P}^1)$. Hence $G(c_2)$ must necessarily be the projection onto $H^2(\mathbf{P}^1)$ and the proof is complete. $\square$