ScalingStacks

Proof. Let $M\in\qc(X)$ be dualizable with dual $M^{\vee}$. Then for any map $\eta:\Spec A\rightarrow X$, the pullback $\eta^{*}M$ is dualizable with dual $\eta^{*}M^{\vee}$. Dualizable objects of $\Mod_{A}$ are perfect, hence $\eta^{*}M$ is perfect and so by definition, $M$ is perfect.

Now suppose $M\in\qc(X)$ is perfect. Recall that by definition, we have

$$\qc(X)\simeq\lim_{\Spec A\in{\it Aff}/X}\Mod_{A}.$$

Since $M$ is perfect, for any map $\eta:\Spec A\to X$, the pullback $\eta^{*}M$ is perfect, hence dualizable. We take the value of the dual $M^{\vee}$ along a map $\eta:\Spec A\to X$ to be the dual of the pullback $(\eta^{*}M)^{\vee}$. Note that $M^{\vee}$ is well-defined, since for any composite $\eta\circ\nu:\Spec B\rightarrow X$, there is a natural equivalence $(\nu^{*}\eta^{*}M)^{\vee}\simeq\nu^{*}((\eta^{*}M)^{\vee})$.

To exhibit $M$ and $M^{\vee}$ as dual to one another, we must construct the requisite unit and counit maps $u:\mathcal{O}_{X}\rightarrow M\otimes M^{\vee}$ and $c:M^{\vee}\otimes M\rightarrow\mathcal{O}_{X}$. Again using the definition of $\qc(X)$ as a limit, to produce one of these maps, it suffices to define analogous maps for the pullbacks under each $\eta:\Spec A\rightarrow X$ which themselves are compatible under pullbacks. But the existence of such maps are an immediate consequence of the definition $\eta^{*}M^{\vee}=(\eta^{*}M)^{\vee}$. Finally, to verify that the usual composititions $M\rightarrow M\otimes M^{\vee}\otimes M\rightarrow M$ and $M^{\vee}\rightarrow M^{\vee}\otimes M\otimes M^{\vee}\rightarrow M^{\vee}$ are equivalences, it suffices to check under pullbacks to affines. But this is a direct consequence of our definition of $M^{\vee}$ and the fact that pullbacks preserve tensor products. ∎