This question is related to the paper "Frobenius algebras and ambidextrous adjunctions" by Aaron Lauda (https://arxiv.org/abs/math/0502550). Below $\Sigma\mathrm{Vect}$ is the one-object delooping of vector spaces and I will only talk about multiplication, but everything applies to comultiplication as well.
In section 2 there is a nice description of how Frobenius objects in a category $K$ come from an ambidextrous adjunction in $EM(K)$, the Eilenberg-Moore completion. It turns out that $K$ embeds into $EM(K)$.
In particular, Corollary 21 says that every 2D TQFT arises from an ambidextrous adjunction in $EM(\Sigma\mathrm{Vect})$, which has algebras as objects, bimodules as morphisms and bimodule homomorphisms as $2$-morphisms and the implication goes in the other direction as well, i.e. if we have an ambijunction, then we get a Frobenius object.
However, a TQFT valued in $\mathrm{Vect}$ corresponds to a (co)commutative Frobenius algebra. This is the picture I have in mind:
Let's say $A$ is a Frobenius algebra. In $EM(\Sigma\mathrm{Vect})$ we have two objects $B, B^A$ where $B$ is the (image of the) object of $\Sigma\mathrm{Vect}$ and we can write $A = UF$ with $$U:B^A\to B, F:B\to B^A$$ left-right adjoints of each other. The multiplication looks like $$UFUF\to UF$$ where we use the counit map $FU\to 1$ in the middle - so, we use $F$ from the left factor and $U$ from the right factor. Comultiplication, unit, counit maps have similar descriptions.
What I would like is an example of these $B^A, U, F$ where $A$ is some simple (co)commutative Frobenius algebra. To be honest, I don't know which algebra $B$ should be either, but I believe it should be $k$, the underlying field.
To talk about commutativity, we would need a symmetric monoidal structure on $\operatorname{Hom}(B, B)$, i.e., on $(B, B)$-bimodules, but I fail to see how the asymmetry in using $F$ from the left factor and $U$ from the right goes away. Examples of similar constructions in other categories (or references for such examples) are also welcome. Thanks.
