**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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