Lemma 15.126.3. Let $R$ be a ring. Let $M$ be an object of $D(R)$. The following are equivalent

1. $M$ has a left dual in $D(R)$ as in Categories, Definition 4.43.5,

2. $M$ is a perfect object of $D(R)$.

Moreover, in this case the left dual of $M$ is the object $M^\vee$ of Lemma 15.74.15.

Proof. If $M$ is perfect, then we can represent $M$ by a bounded complex $M^\bullet$ of finite projective $R$-modules. In this case $M^\bullet$ has a left dual in the category of complexes by Lemma 15.72.2 which is a fortiori a left dual in $D(R)$.

Assume (1). Say $N$, $\eta : R \to M \otimes _ R^\mathbf {L} N$, and $\epsilon : M \otimes _ R^\mathbf {L} N \to R$ is a left dual as in Categories, Definition 4.43.5. Choose a complex $M^\bullet$ representing $M$. Choose a K-flat complexes $N^\bullet$ with flat terms representing $N$, see Lemma 15.59.10. Then $\eta$ is given by a map of complexes

$\eta : R \longrightarrow \text{Tot}(M^\bullet \otimes _ R N^\bullet )$

We can write the image of $1$ as a finite sum

$\eta (1) = \sum \nolimits _ n \sum \nolimits _ i m_{n, i} \otimes n_{-n, i}$

with $m_{n, i} \in M^ n$ and $n_{-n, i} \in N^{-n}$. Let $K^\bullet \subset M^\bullet$ be the subcomplex generated by all the elements $m_{n, i}$ and $\text{d}(m_{n, i})$. By our choice of $N^\bullet$ we find that $\text{Tot}(K^\bullet \otimes _ R N^\bullet ) \subset \text{Tot}(M^\bullet \otimes _ R N^\bullet )$ and $\eta (1)$ is in the subcomplex by our choice above. Denote $K$ the object of $D(R)$ represented by $K^\bullet$. Then we see that $\eta$ factors over a map $\tilde\eta : R \longrightarrow K \otimes _ R^\mathbf {L} N$. Since $(1 \otimes \epsilon ) \circ (\eta \otimes 1) = \text{id}_ M$ we conclude that the identity on $M$ factors through $K$ by the commutative diagram

$\xymatrix{ M \ar[rr]_-{\eta \otimes 1} \ar[rrd]_{\tilde\eta \otimes 1} & & M \otimes _ R^\mathbf {L} N \otimes _ R^\mathbf {L} M \ar[r]_-{1 \otimes \epsilon } & M \\ & & K \otimes _ R^\mathbf {L} N \otimes _ R^\mathbf {L} M \ar[u] \ar[r]^-{1 \otimes \epsilon } & K \ar[u] }$

Since $K$ is bounded above it follows that $M \in D^-(R)$. Thus we can represent $M$ by a bounded above complex $M^\bullet$ of free $R$-modules, see for example Derived Categories, Lemma 13.15.4. Write $\eta (1) = \sum \nolimits _ n \sum \nolimits _ i m_{n, i} \otimes n_{-n, i}$ as before. By Lemma 15.126.2 we can find a subcomplex $K^\bullet \subset M^\bullet$ containing all the elements $m_{n, i}$ which is bounded and consists of finite free $R$-modules. As above we find that the identity on $M$ factors through $K$. Since $K$ is perfect we conclude $M$ is perfect too, see Lemma 15.74.5. $\square$

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