Normalized trace and independence of a projective embedding

The adjunction [F2] identifies $R{\mathrm{Hom}}_X(K,D_i)$ with $R{\mathrm{Hom}}_P(i_{\ast }K,{\omega }_P[N])$. The pushforward $i_{\ast }K$ is bounded coherent, so [F3] identifies the latter with $R{\mathrm{Hom}}_k(R\mathrm{\Gamma }(P,i_{\ast }K),k)$, which equals $R{\mathrm{Hom}}_k(R\mathrm{\Gamma }(X,K),k)$ by [F2]. Cohomology in degree $r$ gives the displayed formula. The comparison is the evaluation pairing: for maps $\alpha :K\to D_i[r]$ and $\beta :{\mathcal{O}}_X\to K[-r]$, adjunction takes $\alpha$ to the ambient map followed by the counit. It takes $\beta$ followed by $i_{\ast }{\mathcal{O}}_X$ to the same ambient composition; pulling back the unit ${\mathcal{O}}_P\to i_{\ast }{\mathcal{O}}_X$ makes the equality immediate by evaluation at $1$. Thus the paired scalar is $t_i(\alpha [-r]\circ \beta )$, not merely an unspecified vector-space isomorphism.

At $r=0$, each embedding gives a representation on $D_{\mathrm{C}\mathrm{o}\mathrm{h}}^b(X)$ of the same contravariant functor $K\mapsto {\mathrm{Hom}}_k(H^0(X,K),k)$. Both representing objects lie in that category by [F1]. The Yoneda evaluation bijection [F3] supplies a unique isomorphism: evaluate the natural comparison at the first representing object on its identity, and do the reverse at the second; naturality says that the two composites are identities. The trace itself is recovered by evaluating the represented functional for a map ${\mathcal{O}}_X\to D_i$ at $1\in H^0(X,{\mathcal{O}}_X)$, so this isomorphism preserves it. Naturality under shifts then preserves every degree and triangle comparison. These unique isomorphisms compose transitively, proving independence of the embedding with its normalization; an unnormalized dualizing complex alone is not claimed to be uniquely isomorphic. AC is retained from [F1]–[F3].