Recall that an object $a$ in a symmetric monoidal category $(\mathcal{C}, \otimes, e)$
is *dualizable* if there exists an object $b$ and morphisms $\varepsilon\colon b \otimes a \to e$
and $\eta\colon e \to a \otimes b$ such that the compositions
$$
a \xrightarrow{\eta \otimes a} a \otimes b \otimes a \xrightarrow{a\otimes \varepsilon} a
$$
and
$$
b \xrightarrow{b \otimes \eta} b \otimes a \otimes b \xrightarrow{\varepsilon \otimes b} b
$$
are the identities.

Let $\mathcal{O}$ be a ring in a topos $\mathcal{T}$.
An object in the category of $\mathcal{O}$-modules is dualizable
if and only if it is locally isomorphic to a direct summand of a finite free module.
A *strictly perfect complex* is a bounded cochain complex of direct summands of finite free modules.
An object in the category of cochain complexes of $\mathcal{O}$-modules is dualizable
if and only if it is locally isomorphic to a strictly perfect complex.
For instance, if $\mathcal{T}$ is the category of sets,
then the dualizable objects in the categories mentioned above are the finite projective modules
and the bounded complexes of finite projective modules, respectively.

A complex in the derived category $D(\mathcal{O})$ of $\mathcal{O}$-modules is called *perfect*
if it is locally quasi-isomorphic to a strictly perfect complex.
Any perfect complex in $D(\mathcal{O})$ is dualizable.

**My question is: Is the converse statement true?
That is, is any dualizable object in $D(\mathcal{O})$ perfect?**

I know that this is true in the special case when $\mathcal{T}$ is the category of sets. One argument for this is that $\mathcal{O}$ is compact in $D(\mathcal{O})$, something which is not true in any topos, from which it easily follows that any dualizable complex is compact. Hence the statement follows from the fact that compact objects are perfect, again using that $\mathcal{T}$ is the category of sets.