Lemma 45.9.9. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). If $X$ and $Y$ are smooth projective schemes over $k$, then $H^*(X \amalg Y) \to H^*(X) \times H^*(Y)$, $a \mapsto (i^*a, j^*a)$ is an isomorphism where $i$, $j$ are the coprojections.

Proof. If $X$ or $Y$ is empty, then this is true because $H^*(\emptyset ) = 0$ by Lemma 45.9.5. Thus we may assume both $X$ and $Y$ are nonempty.

We first show that the map is injective. First, observe that we can find morphisms $X' \to X$ and $Y' \to Y$ of smooth projective schemes so that $X'$ and $Y'$ are equidimensional of the same dimension and such that $X' \to X$ and $Y' \to Y$ each have a section. Namely, decompose $X = \coprod X_ d$ and $Y = \coprod Y_ e$ into open and closed subschemes equidimensional of dimension $d$ and $e$. Then take $X' = \coprod X_ d \times \mathbf{P}^{n - d}$ and $Y' = \coprod Y_ e \times \mathbf{P}^{n - e}$ for some $n$ sufficiently large. Thus pullback by $X' \amalg Y' \to X \amalg Y$ is injective (because there is a section) and it suffices to show the injectivity for $X', Y'$ as we do in the next parapgrah.

Let us show the map is injective when $X$ and $Y$ are equidimensional of the same dimension $d$. Observe that $[X \amalg Y] = [X] + [Y]$ in $\mathop{\mathrm{CH}}\nolimits ^0(X \amalg Y)$ and that $[X]$ and $[Y]$ are orthogonal idempotents in $\mathop{\mathrm{CH}}\nolimits ^0(X \amalg Y)$. Thus

$1 = \gamma ([X \amalg Y] = \gamma ([X]) + \gamma ([Y]) = i_*1 + j_*1$

is a decomposition into orthogonal idempotents. Here we have used Lemmas 45.9.5 and 45.9.6 and axiom (C)(c). Then we see that

$a = a \cup 1 = a \cup i_*1 + a \cup j_*1 = i_*(i^*a) + j_*(j^*a)$

by the projection formula (Lemma 45.9.1) and hence the map is injective.

We show the map is surjective. Write $e = \gamma ([X])$ and $f = \gamma ([Y])$ viewed as elements in $H^0(X \amalg Y)$. We have $i^*e = 1$, $i^*f = 0$, $j^*e = 0$, and $j^*f = 1$ by axiom (C)(a). Hence if $i^* : H^*(X \amalg Y) \to H^*(X)$ and $j^* : H^*(X \amalg Y) \to H^*(Y)$ are surjective, then so is $(i^*, j^*)$. Namely, for $a, a' \in H^*(X \amalg Y)$ we have

$(i^*a, j^*a') = (i^*(a \cup e + a' \cup f), j^*(a \cup e + a' \cup f))$

By symmetry it suffices to show $i^* : H^*(X \amalg Y) \to H^*(X)$ is surjective. If there is a morphism $Y \to X$, then there is a morphism $g : X \amalg Y \to X$ with $g \circ i = \text{id}_ X$ and we conclude. To finish the proof, observe that in order to prove $i^*$ is surjective, it suffices to do so after tensoring by a nonzero graded $F$-vector space. Hence by axiom (B)(b) and nonvanishing of cohomology (Lemma 45.9.5) it suffices to prove $i^*$ is surjective after replacing $X$ and $Y$ by $X \times \mathop{\mathrm{Spec}}(k')$ and $Y \times \mathop{\mathrm{Spec}}(k')$ for some finite separable extension $k'/k$. If we choose $k'$ such that there exists a closed point $x \in X$ with $\kappa (x) = k'$ (and this is possible by Varieties, Lemma 33.25.6) then there is a morphism $Y \times \mathop{\mathrm{Spec}}(k') \to X \times \mathop{\mathrm{Spec}}(k')$ and we find that the proof is complete. $\square$

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