# Serre duality in derived category

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 ?