ScalingStacks

Proof of Proposition 4.1. We first prove that $\mathrm{Mod}_{A}(\mathcal{C})\otimes_{\mathcal{C}}\mathcal{M}$ is equivalent to $\mathrm{Mod}_{A}(\mathcal{M})$ by the natural evaluation functor. Consider the adjunction

where $F(-)=A\otimes-$ is the induction, and $G$ is the forgetful functor.

The above adjunction induces an adjunction

and thus a functor to modules over the monad $T=(G\otimes{\rm id})\circ(F\otimes{\rm id})$ acting on $\mathcal{M}$. The functor underlying $T$ is given by tensoring with $A$, so we also have an equivalence $\mathrm{Mod}_{T}(\mathcal{M})\simeq\mathrm{Mod}_{A}(\mathcal{M})$.

By its universal characterization, the functor $G\otimes{\rm id}$ is colimit preserving. Note as well that $G$ and hence $G\otimes{\rm id}$ is also $\mathcal{C}$-linear (or in other words, the adjunction satisfies an analogue of the projection formula). Thus it follows from Lemma 4.3 that $G\otimes{\rm id}$ is also conservative. Thus $G\otimes{\rm id}$ satisfies the monadic Barr-Beck conditions, and we obtain the desired equivalence $\mathrm{Mod}_{A}(\mathcal{C})\otimes_{\mathcal{C}}\mathcal{M}\simeq\mathrm{Mod}_{A}(\mathcal{M})$.

Next, we can apply this to the instance where $\mathcal{M}$ is the $\infty$-category of left modules over another associative algebra $A^{\prime}$ to conclude that there is a natural equivalence $\mathrm{Mod}_{A}(\mathcal{C})\otimes_{\mathcal{C}}\mathrm{Mod}_{A^{\prime}}(\mathcal{C})\simeq\mathrm{Mod}_{A}(\mathrm{Mod}_{A^{\prime}}(\mathcal{C}))$. We now have a chain of adjunctions

in which the composite $G^{\prime}\circ G^{\prime\prime}$ is colimit preserving and conservative, and hence satisfies the monadic Barr-Beck conditions.

Furthermore, the above adjunction naturally extends to a diagram in which the cycle of left adjoints (denoted by bowed arrows), and hence also the cycle of right adjoints (denoted by straight arrows), commute

Here $F(-)=A_{1}\otimes A_{2}\otimes-$ is the induction, $G$ is the forgetful functor, $f$ is the natural functor factoring through $\mathrm{Mod}_{A}(\mathcal{C})\otimes_{\mathcal{C}}\mathrm{Mod}_{A^{\prime}}(\mathcal{C})$, and $g$ is its right adjoint. From this diagram, we obtain a morphism of monads

Now the underlying functors of the monads $GF(-)$ and $G^{\prime}G^{\prime\prime}F^{\prime\prime}F^{\prime}(-)$ are both equivalent to the tensor $A\otimes A^{\prime}\otimes(-)$, so the above morphism of monads is an equivalence. Thus we obtain the promised equivalence $\mathrm{Mod}_{A}(\mathcal{C})\otimes_{\mathcal{C}}\mathrm{Mod}_{A^{\prime}}(\mathcal{C})\simeq\mathrm{Mod}_{A}(\mathrm{Mod}_{A^{\prime}}(\mathcal{C}))\simeq\mathrm{Mod}_{A\otimes A^{\prime}}(\mathcal{C})$.

Finally, we show that the $\infty$-category of left $A$-modules $\mathrm{Mod}_{A}(\mathcal{C})$ is a dualizable $\mathcal{C}$-module by directly exhibiting the $\infty$-category of right $A$-modules $\mathrm{Mod}_{A^{\rm op}}(\mathcal{C})$ as its dual. The trace map is given by the two-sided bar construction

$$\tau:\mathrm{Mod}_{A}(\mathcal{C})\otimes_{\mathcal{C}}\mathrm{Mod}_{A^{\rm op}}(\mathcal{C})\to\mathcal{C}\qquad M,N\mapsto M\otimes_{A}N$$

The unit map is given by the induction

$$u:\mathcal{C}\to\mathrm{Mod}_{A^{\rm op}}(\mathcal{C})\otimes_{\mathcal{C}}\mathrm{Mod}_{A}(\mathcal{C})\simeq\mathrm{Mod}_{A^{\rm op}\otimes A}(\mathcal{C})\qquad c\mapsto A\otimes c$$

where we regard $A\otimes c$ as an $A$-bimodule.

One can verify directly that the composition

is equivalent to the identity. First, $({\rm id}\otimes u)(M)$ is equivalent to $A\otimes M$ regarded as an $A\otimes A^{\rm op}\otimes A$-module, and second, $(\tau\otimes{\rm id})(A\otimes M)$ is equivalent to $A\otimes_{A}M\simeq M$. ∎