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](https://mathoverflow.net/questions/55439/flat-quasi-coherent-modules-as-colimits-of-locally-free-ones) 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.



Dualizable objects are flat?
A dual pair as you describe above induces an adjunction $(X^\*\otimes-)\dashv(X\otimes-)$, and since right adjoints preserve limits, $X$ is flat. 


