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I have trouble understanding a point of Milne's article "Duality in the flat cohomology of a surface"

see the "Alternatively" on p. 177, paragraph before Corollary 1.10:

Let $X/\mathbf{F}_q$ be a smooth proper variety of dimension $m$. Denote $\Omega_{X/\mathbf{F}_q}^i$ by $\Omega^i$.

  1. Why does it follow from Grothendieck duality that $H^i(\Omega_{d=0}^r) \to H^{m-i}(\Omega^{m-r}/d\Omega^{m-r+1})^*$ is an isomorphism? Here, the map is induced from the pairing (see Lemma 1.8) $\Omega_{d=0}^r \times \Omega^{m-r}/d\Omega^{m-r+1} \to \Omega^m, (\omega,\tau) \mapsto C(\omega \wedge \tau)$ with $C$ the Cartier operator. [The analogous statement for $\Omega^i \times \Omega^{m-i} \to \Omega^m$ is clear to me.]
  2. Why does the theorem follow from "the usual formalism" (It is clear to me that the two rows on the top of p. 177 are exact (long exact cohomology sequence for the complexes $X^\cdot$ and $Y^\cdot$), but not that the ladder commutes.)?

It seems that one has to exploit the pairing $X^\cdot \otimes Y^\cdot \to Z^\cdot$, see Lemma 1.8.

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Have you tried looking at the exposition of the theorem in: Berthelot, P. Le théorème de dualité plate pour les surfaces (d'après J. S. Milne). (French) [The flat duality theorem for surfaces (according to J. S. Milne)] Algebraic surfaces (Orsay, 1976--78), pp. 203--237, Lecture Notes in Math., 868, Springer, Berlin-New York, 1981. MR0638601? – anon Dec 20 '11 at 10:03
have you asked james? – roy smith Dec 26 '11 at 5:09
Yes. He didn't remember his argument. – Timo Keller Jan 3 '12 at 17:12

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