Let me first recall some basic well-known definitions: Let $R$ be a ring (as always commutative). A (cocomplete) *abelian tensor category* is defined to be a symmetric monoidal category, whose underlying category is also a $R$-linear (cocomplete) abelian category, such that the tensor product is right exact (cocontinuous) and $R$-linear in each variable. An object $X$ of such a category $C$ is called *flat* if $X \otimes -$ is an exact functor. Also recall that an object $X$ of a symmetric monoidal category is called *dualizable* with dual $X^\*$ if it comes equipped with morphisms $1_C \to X \otimes X^\*$ and $X^\* \otimes X \to 1_C$ satisfying the two triangular identities.

Now these notions are connected in the case of $C = \text{Mod}(R)$: The dualizable objects are precisely the finitely generated projective $R$-modules (equivalently, locally free of finite rank). Thus, every dualizable object is flat, and Lazard's Theorem implies that every flat module is a directed colimit of locally free modules. Now I wonder if these statements remain true in the general case. I've already this concerning Lazard's theorem here for the category of quasi-coherent modules on a nice scheme (without answer). Thus my question is:

**Question:** Is every dualizable object in a (cocomplete) abelian tensor category flat?

Remark that we cannot insert "projective" as an intermediate step. It is easy to see that $1$ is projective iff every dualizable object is projective. But $1$ is not projective in $\text{Qcoh}(S)$ forr some non-affine concentrated scheme $S$, whereas the question in this case is answered with "yes" since every locally free quasi-coherent sheaf is locally flat and thus (globally) flat.