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$\newcommand{\Hom}{\mathscr{Hom}}$ Let $C$ be a cocomplete closed symmetric monoidal category, and the tensor product preserves colimits in each variable;

Let $A$ be a commutative algebra in $C$, together with a morphism $p: A\to 1$ such that the composite $A\otimes A\overset\mu\to A\overset p\to 1$ is a perfect pairing, i.e. it induces an isomorphism $A\to \Hom(A,1)$, where $\Hom$ denotes the internal hom.

Q1 Is it enough to conclude that $A$ has a structure of a Frobenius algebra in $C$ ? Where the comultiplication would be obtained by dualizing the multiplication ?

If I understand correctly the remark here, they seem to claim it's true in the category of vector spaces over a field - is it true more generally ? (this also seems to be definition 3.1 in the paper of Ross Street that is linked there, but because it talks about pseudomonoids in 2-categories etc. it's not clear that this really refers to the same thing)

My second question is the main one, I'm interested in it even if the answer to Q1 is no. I'm interested about the traces of $A$-modules. Indeed, it's not hard to show that an $A$-module which is dualizable is also dualizable as an object in $C$. In particular, it has a trace as an $A$-module (which is an element in $\hom_A(A,A) = \hom(1,A)$), and as an object of $C$ (an element in $\hom(1,1)$). My question is :

Q2 : can these traces be compared in any way ? For instance, is it true that $Tr(A) p\circ Tr_A(x) \circ \eta = Tr(x)$ ? What about traces of $A$-module morphisms $x\to x$ ?

(where $\eta : 1\to A$ is the structure map, and $Tr_A$ is the trace as an $A$-module, $Tr$ the trace in $C$)

If a result holds under additional niceness assumptions, I'd be happy to hear about them too.

The reason I believe this isn't too far-fetched is that under these assumptions, $\Hom(A,1)\cong A$ as $A$-modules ($\Hom$ is the internal hom), and from that isomorphism it follows that $\Hom_A(x,A) \cong \Hom(x,1)$ for any $A$-module $x$, i.e. the dual is the same as an $A$-module and as an object in $C$. If you draw the "obvious" diagrams relating $Tr_A(x)$ and $Tr(x)$, some of them commute, and some others I'm not sure, but they seem to indicate that a relation as above could hold.

Moreover, here's a related paper, in which the authors show that a Frobenius functor preserves duality data in a sense, and they claim that $A\otimes - : C\to C$ is Frobenius when $A$ is a Frobenius ring. What I'm wondering is, in a sense, whether the forgetful $A-Mod \to C$ is also a Frobenius functor, so let me state this as a 3rd question, although I'm only interested in it insofar as it could help for Q2:

Q3: Is the forgetful functor $A-Mod\to C$ a Frobenius functor in the sense of the above paper ?

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Let me write an answer, to at least provide the correct formula, even though I'm not completely satisfied with it. The reason I'm not satisfied is that I secretly want to make this argument in a symmetric monoidal $(\infty,2)$-category, and I want the calculation to be natural in dualizability data, and I don't know enough about symmetric monoidal $(\infty,2)$-categories to know what parts of the argument I need to adapt to make it work.

With that in mind, here's the proof.

The answer to Q1 is yes, provided I assume $A$ is dualizable [which is an assumption I'm willing to make]

In the case of vector spaces, the assumption that $A\cong A^*$ is enough to impose dualizability anyway, but I'm not sure it is the case in general, so let me add that hypothesis. I don't know what the original reference for this fact would be, so let me just point out that it is worked out in these slides by Andrew Baker, where the coproduct structure is defined in terms of the co-evaluation $k\to A\otimes A$ and the dual pairing $A\otimes A \to k$ (in my case given by the composite $A\otimes A\to A\to k$). Once the co-evaluation has been introduced (the slides say it exists "by standard linear algebra", but in my case I assume it does), all the constructions and proofs are diagrammatic, and so they work in any symmetric monoidal category.

To answer Q2, let me first answer Q3, to which the answer is also yes. The point is that a Frobenius algebra is in particular a co-algebra, with $A\to A\otimes A$ an $A$-bimodule morphism. In particular, the forgetful functor (which is always lax symmetric monoidal) can be given an oplax symmetric monoidal structure with structure maps $x\otimes_A y = x\otimes_A A \otimes_A y \to x\otimes_A A\otimes A\otimes_A y \cong x\otimes y$. If I'm not mistaken, this can be shown to be an oplax structure, and one can also show the Frobenius compatibility conditions from the paper mentioned in the question

(I haven't checked that they were the same, but there's another approach to this oplax structure : the free $A$-module functor, $x\mapsto A\otimes x$, is left adjoint to the forgetful functor. But, our assumptions imply that $A$ is isomorphic, as an $A$-module, to $A^*$, so that this functor is isomorphic to $\mathscr{Hom}(A,x)$, i.e. the right adjoint to the forgetful functor - in particular, this gives the forgetful functor an oplax symmetric monoidal structure)

With this in mind, we can simply apply the result of the paper mentioned in the question, to see that if $(x,y, \eta,\epsilon)$ is a dualizability datum in $A-Mod$, then a certain modification of it is a dualizability datum in $C$. If you look at what this modification is, you'll get by an immediate diagram chase:

The answer to Q2 is that if we let $e: 1\to A$ denote the composite $1\to A\otimes A\to A$, where $1\to A\otimes A$ is the co-evaluation map coming from the self-dualizability of $A$, then $p\circ (e \cdot Tr_A(x)) \circ h = Tr(x)$, where $h:1\to A$ is the unit of the algebra structure.

Now, the thing is that we don't really need the Frobenius functor formalism to make that last claim, indeed if you look at the paper in question, the claim for dualizability data is just an easy computation. In particular, it seems as though there's no need for all the higher coherence structure to make that claim, and it looks like you can just say "this is a functor from dualizability data to dualizability data".

So it seems like this result should extend to a higher coherent setting, such as symmetric monoidal $(\infty,2)$-categories, by replacing $=$ with "isomorphic, naturally in the dualizability data", but I'm not sure how much of the theory has been set up in this setting, so I would be very grateful for any references on the matter.

In particular, I think I'm looking for a claim of the form

Let $C^{dual}$ denote the subcategory of $C$ on dualizable objects and dualizable morphisms, and $C^{dual-data}$ the category of dualizability data and dualizability data morphisms. Then the forgetful functor $C^{dual-data}\to C^{dual}$ is an equivalence.

This is true if $C$ is a symmetric monoidal $(\infty,1)$-category, and can certainly be found somewhere in Higher Algebra (specifically, lemma 4.6.1.10.), and I think it's true in $(\infty,2)$-categories too but I don't know the references too well. Note that the difference would be that there are noninvertible dualizability data morphisms in that setting.

The other type of claim I'm missing, is whether I can make the description in the proof of Theorem 2 of the paper I mentioned in my question functorial in the dualizability data, in the $(\infty,2)$-categorical sense.

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