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Let $X$ be a smooth projective variety over a field $k$, and $\omega_X$ its canonical bundle. Denote by $D^b(X):=D^b(\mathbf{Coh})$ the bounded derived category of coherent sheaves on $X$.

$\textbf{Theorem}$ : For any two comples $\mathcal{E}^\bullet,\mathcal{F}^\bullet$ in $D^b(X)$ there exists a functorial isomorphism $$\eta : Ext^i(\mathcal{E}^\bullet,\mathcal{F}^\bullet) \to Ext^{n-i}(\mathcal{F}^\bullet,\mathcal{E}^\bullet\otimes \omega_X)^*,$$ where $^*$ denotes the vector space dual.

This theorem can be found in D. Huybrechts's $\textit{Fourier-Mukai Transforms in Algebraic Geometry}$ (theorem 3.12). It basically says that the exact functor $$\cdot\otimes\omega_X[n] : D^b(X) \to D^b(X)$$ is a Serre functor in the sense of Bondal and Krapanov (see here for instance).

Now my question is: how can we prove this theorem? I know that this is a particular case of Grothendieck-Verdier duality, but is it possible to avoid it?

$\textbf{My attempts:}$ I know the classical Serre Duality (Harthorne's $\textit{Algebraic Geometry}$, III.7), which in particular says that the theorem is true when $\mathcal{E}^\bullet = \mathcal{E}$ is a locally free sheaf and $\mathcal{F}^\bullet=\mathcal{F}$ is any coherent sheaf. For the general case, I tried to use the definition: $$Ext^i(\mathcal{E}^\bullet,\mathcal{F}^\bullet) = H^i(RHom^\bullet(\mathcal{E}^\bullet,\mathcal{F}^\bullet)),$$ where $Hom^n(A^\bullet,B^\bullet)=\bigoplus Hom(A^k,B^{k+n})$ with $d(f)=d_B\circ f - (-1)^nf\circ d_A$. I tried to replace $\mathcal{F}^\bullet$ by a quasi-isomorphic complex of injectives (of quasi-coherent sheaves) and $\mathcal{E}$ by a quasi-isomorphic complex of locally free sheaves so I can use the classical Serre duality on each summand of the direct sum, but I'm not sure this is a way to start. If someone could give me some help I would really appreciate.

$\textbf{EDIT} :$ Thanks to the answer of @mayer_vietoris, I tried the following. Suppose $\mathcal{E}^\bullet$ is a complex of locally free sheaves and $\mathcal{F}$ is a complex of injectives so that $RHom^\bullet(\mathcal{E}^\bullet,\mathcal{F}^\bullet)=Hom^\bullet(\mathcal{E}^\bullet,\mathcal{F}^\bullet).$ $$\begin{eqnarray} Hom^i(\mathcal{E}^\bullet,\mathcal{F}^\bullet) &=& \bigoplus Hom(\mathcal{E}^k,\mathcal{F}^{k+i}), \\ &=& \bigoplus Ext^0(\mathcal{O}_X,(\mathcal{E}^k)^ \vee\otimes\mathcal{F}^{k+i}),\\ &\simeq& \bigoplus Ext^n(\mathcal{F}^{k+i},\mathcal{E}^k\otimes\omega_X)^*, \\ &\simeq& \bigoplus Hom(\mathcal{F}^{k+i},\mathcal{E}^k\otimes\omega_X[n])^*,\\ &\simeq& Hom^{n-i}(\mathcal{F}^\bullet,\mathcal{E}^\bullet\otimes\omega_X)^*. \end{eqnarray}$$ Up to a replacement of $\mathcal{E}^\bullet\otimes\omega_X$ by a complex of injectives, I would like to conclude by replacing $Hom^i(\dots$ by $H^i(Hom^\bullet(\dots$ and similarly with $Hom^{n-i}$. Thus I obtain the desired equality. Am I wrong somewhere ?

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I believe that you can find out what you look for explicitly in a very beautiful answer here, The idea behind the notion of dualizing sheaf.

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  • $\begingroup$ Thank you for your answer. Indeed the answer given in the other post is really interesting, but it does not deal with complexes (I mean, it only treats complexes concentrated in degree 0, i.e. coherent sheaves). $\endgroup$ – Dominique MATTEI May 1 '18 at 16:21

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