I am dealing with monoidal bicategory which has algebras, bimodules and bimodule maps as objects, 1-morphisms and 2-morphisms respectively.
Let $k$ be a commutative ring. We take any $k$-algebra $A$ and claim that its dual is the algebra $A^{op}$. The algebra morphisms are taken in the form of bimodules. In particular, $id_A = {}_AA_A$, $id_{A^{op}} = {}_{A^{op}}A_{A^{op}}$, $ev_A = {}_{A \otimes_k A^{op}}A_k$ and $coev_A = {}_k A_{A^{op} \otimes_k A}$. Composition of bimdolues is given by $\otimes_?$ where "?" denotes intermediate algebra. My attempt is:
$$(ev_A \otimes_k id_A) \otimes_? (id_A \otimes_k coev_A)$$ $$= (ev_A \otimes_? id_A) \otimes_k (id_A \otimes_? coev_A)$$ $$= ({}_{A \otimes_k A^{op}}A_k \otimes_? {}_AA_A) \otimes_k ({}_AA_A \otimes_? {}_kA_{A^{op} \otimes_k A})$$ $$= ({}_AA_A \otimes_? {}_AA_A) \otimes_k ({}_AA_A \otimes_? {}_AA_A)$$ $$= ( {}_AA_A) \otimes_k ({}_AA_A)$$ $$= {}_AA_A$$
where in the 2nd line I use $(f \otimes g) \circ (f' \otimes g') = (f \circ f') \otimes (g \circ g')$, in the 4th line I substitute both of ${}_{A \otimes_k A^{op}}A_k$ and ${}_kA_{A^{op} \otimes_k A}$ with ${}_AA_A$ following this equality and isomorphism respectively (can I just do this? why? clearly its important what we view the bimodule as, since the two sides represent the source and target object), and in the 5th line I used the fact that $ {}_AA_A$ is the identity in terms of composition $\otimes_?$ for A-bimodule, and in the last line I used that monoidal tensor of the two objects which are the same will give the same object
which is confusing as I was looking at it as a bimodule until now which is the 1-morphism, and in this step as the algebra which is the 0-morphism/object; although it would make sense because the tensor of same 1-morphisms will act on the tensor of same source objects which in turn give just the same source object automatically thus we just get the same 1-morphism. That is, monoidal tensor of two same 1-morphisms is the same 1-morphism (?)
Is this correct so far? If so, does this successfully prove the zigzag equation and thus the claim?
I feel I would be more comfortable dealing with algebra morphisms directly than bimodules, although what would the $ev_A, coev_A$ maps be explicitly then as just morphisms? Or more generally, what would be the "recipe" for converting between an explicit morphism and the associated bimodule (and vice versa)? The particular case presented here would be a nice expository example for me.