Lemma 15.69.14. 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 $K = (K^\vee )^\vee$. There are functorial isomorphisms

$K^\vee \otimes _ A^\mathbf {L} L = R\mathop{\mathrm{Hom}}\nolimits _ A(K, L) \quad \text{and}\quad H^0(K^\vee \otimes _ A^\mathbf {L} L) = \mathop{\mathrm{Ext}}\nolimits _ A^0(K, L)$

for $L \in D(A)$.

Proof. We can represent $K$ by a complex $K^\bullet$ of finite projective $A$-modules. By Lemma 15.68.2 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 $K^\bullet$. Thus the formula $(K^\vee )^\vee = K$ is clear from the fact that the double dual of a finite projective module is itself.

The second equality follows from the first by Lemma 15.68.1 and Derived Categories, Lemma 13.19.8 as well as the definition of Ext groups, see Derived Categories, Section 13.27. Let us prove the first equality.

Let $L^\bullet$ be a complex of $A$-modules representing $L$. The object on the left of the first equality is represented by $\text{Tot}(E^\bullet \otimes _ A L^\bullet )$. The object on the right of the first equality sign is represented by the complex $\mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , L^\bullet )$ by the same lemma as before. Thus the equality follows from the fact that

$\mathop{\mathrm{Hom}}\nolimits _ A(K^ n, A) \otimes _ A L^ m = \mathop{\mathrm{Hom}}\nolimits _ A(K^ n, L^ m)$

for all $n, m$ because $K^ n$ is finite projective. To be a bit more precise we define the map on the level of complexes

$\text{Tot}(E^\bullet \otimes _ A L^\bullet ) = \text{Tot}(\mathop{\mathrm{Hom}}\nolimits ^\bullet (A, L^\bullet ) \otimes _ A \mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , A)) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , L^\bullet )$

using Lemma 15.67.2 and then the statement above shows this is an isomorphism of complexes. $\square$

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