ScalingStacks

Remark 4.9. The above argument also shows that $\qc(X)^{c}$ is dualizable over perfect $k$-modules $\qc(\Spec k)^{c}$ when $X$ is smooth and proper: smoothness is required for the unit in $\qc(X\times X)$ to be perfect and properness is required for the trace to land in perfect $k$-modules.