Derived coherent duality on projective space

By the cohomology calculation [F1], $R\mathrm{\Gamma }(P,{\omega }_P[N])$ has $k$ as its sole cohomology, in degree zero; the Laurent trace identifies it with $k$. Every coherent complex on $P$ is perfect: [F2] gives this for a sheaf, and finite truncation triangles give it for bounded coherent cohomology. Thus derived internal Hom and evaluation are computed locally by bounded finite locally free complexes. Compose the derived cup map and evaluation with the trace to obtain $R\mathrm{\Gamma }(\mathcal{R}\text{\hspace{0.17em}⁣}Hom(K,{\omega }_P[N])){\otimes }_k^{\mathbf{L}}R\mathrm{\Gamma }(K)\to k$. The tensor-Hom adjoint is the map in the statement. It is natural and respects triangles because it is induced by chain evaluation and the signed total-complex differential.

For $K={\mathcal{O}}_P(m)$, the map on degree $a$ cohomology is precisely $H^{N+a}(P,\mathcal{O}(-m-N-1))\to H^{-a}(P,\mathcal{O}(m){)}^{\vee }$, which is an isomorphism in all degrees by [F1], including the zero groups outside their ranges. Hence the map is a quasi-isomorphism for twists and their finite direct sums and shifts. Both functors in the assertion take triangles to triangles contravariantly; exact duality of vector spaces makes the right-hand cohomology equal to the dual of the opposite-degree cohomology. The long exact sequences of [F3] and the five lemma [F3] extend the isomorphism across a cone. Apply this finitely many times to a resolution in [F2], then to truncation triangles of a bounded coherent complex. This proves the assertion for every $K$ and supplies compatibility with connecting maps. AC enters through [F2] and the derived-category and product suppliers.