ScalingStacks

Proof. First, note that the free module $k$, which is the monoidal unit, is clearly compact. Hence all dualizable objects are compact. Moreover, we can write any object as a colimit of free modules. For $M$ compact, the identity map ${\rm id}_{M}\in\Hom(M,M)$ has to factor through a finite colimit, showing that $M$ is perfect. Finally, perfect modules are dualizable since we can explicitly exhibit their dual as a finite limit of free modules. ∎