Lemma 15.74.15. Let $A$ be a ring. Let $K \in D(A)$ be perfect. Then $K^\vee = R\mathop{\mathrm{Hom}}\nolimits _ A(K, A)$ is a perfect complex and $K \cong (K^\vee )^\vee $. There are functorial isomorphisms
for $L \in D(A)$.
Proof. We can represent $K$ by a complex $K^\bullet $ of finite projective $A$-modules. By Lemma 15.73.2 the object $K^\vee $ is represented by the complex $E^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , A)$. Note that $E^ n = \mathop{\mathrm{Hom}}\nolimits _ A(K^{-n}, A)$ and the differentials of $E^\bullet $ are the transpose of the differentials of $K^\bullet $ up to sign. Observe that $E^\bullet $ is the left dual of $K^\bullet $ in the symmetric monoidal category of complexes of $R$-modules, see Lemma 15.72.2. There is a canonical map
which up to sign uses the evaluation map in each degree, see Lemma 15.71.6. (For sign rules see Section 15.72.) Thus this map defines a canonical isomorphism $(K^\vee )^\vee \cong K$ as the double dual of a finite projective module is itself.
The second equality follows from the first by Lemma 15.73.1 and Derived Categories, Lemma 13.19.8 as well as the definition of Ext groups, see Derived Categories, Section 13.27. Let $L^\bullet $ be a complex of $A$-modules representing $L$. By Section 15.72 item (11) there is a canonical isomorphism
of complexes of $A$-modules. This proves the first displayed equality and the proof is complete. $\square$
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