Lemma 45.14.5. Assume given data (D0), (D1), and (D2') satisfying axioms (A1) – (A7). Then axiom (A) of Section 45.9 holds with $\int _ X = \lambda $ as in axiom (A6).

**Proof.**
Let $X$ be a nonempty smooth projective scheme over $k$ which is equidimensional of dimension $d$. We will show that the graded $F$-vector space $H^*(X)(d)[2d]$ is a left dual to $H^*(X)$. This will prove what we want by Homology, Lemma 12.17.5. We are going to use axiom (A5) which in particular says that

Define a map

by multiplying by $\gamma ([\Delta ]) \in H^{2d}(X \times X)(d)$. On the other hand, define a map

by first using pullback $\Delta ^*$ by the diagonal morphism $\Delta : X \to X \times X$ and then using the $F$-linear map $\lambda : H^{2d}(X)(d) \to F$ of axiom (A6) precomposed by the projection $H^*(X)(d) \to H^{2d}(X)(d)$. In order to show that $H^*(X)(d)$ is a left dual to $H^*(X)$ we have to show that the composition of the maps

and

is the identity. If $a \in H^*(X)$ then we see that the composition maps $a$ to

where $q_ i : X \times X \times X \to X$ and $q_{ij} : X \times X \times X \to X \times X$ are the projections, $\Delta _{23} : X \times X \to X \times X \times X$ is the diagonal, and $p_ i : X \times X \to X$ are the projections. The equality holds because $\Delta _{23}^*(q_{12}^*\gamma ([\Delta ]) = \Delta _{23}^*\gamma ([\Delta \times X]) = \gamma ([\Delta ])$ and because $\Delta _{23}^* q_3^*a = p_2^*a$. Since $\gamma ([\Delta ]) \cup p_1^*a = \gamma ([\Delta ]) \cup p_2^*a$ (see below) the above simplifies to

by our choice of $\lambda $ as desired. The second condition $(\epsilon \otimes 1) \circ (1 \otimes \eta ) = \text{id}$ of Categories, Definition 4.43.5 is proved in exactly the same manner.

Note that $p_1^*a$ and $\text{pr}_2^*a$ restrict to the same cohomology class on $\Delta \subset X \times X$. Moreover we have $\mathcal{C}_{\Delta /X \times X} = \Omega ^1_\Delta $ which is the restriction of $p_1^*\Omega ^1_ X$. Hence Lemma 45.14.4 implies $\gamma ([\Delta ]) \cup p_1^*a = \gamma ([\Delta ]) \cup p_2^*a$ and the proof is complete. $\square$

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