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

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

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

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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