ScalingStacks

Proof. The necessity of $M\otimes-$ preserving limits is noted above ($\mathcal{C}$ is closed by virtue of being presentable with monoidal structure distributing over colimits, see [L4, Proposition 2.1.12]). To demonstrate sufficiency, assume that $M\otimes-$ preserves limits, and then consider the endofunctor of $\mathcal{C}$ defined by tensoring with $M$. By assumption on $M$ and $\mathcal{C}$, this functor preserves all limits and colimits. We may now apply the adjoint functor theorem of [L2] to deduce the existence of a left adjoint $F$ to $M\otimes-$. Denote by $M^{\vee}$ the value $F(1_{\mathcal{C}})$ of $F$ applied to the unit of $\mathcal{C}$. The existence of unit and trace maps $1_{\mathcal{C}}\rightarrow M\otimes M^{\vee}\rightarrow 1_{\mathcal{C}}$ is now a particular instance of the unit and counit maps for this adjunction, which implies that $M$ and $M^{\vee}$ are in duality. Hence $M$ is dualizable.

∎

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. ∎

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

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. ∎