What is the composition in SesquiAlg?

To motivate my question, I will describe a famous 2-category. First, there is the 1-category $\text{Vect}$ of vector spaces (over some fixed field). This category has a tensor product $\otimes$, and a notion of algebra: a vector space $A$ and a linear map $m: A \otimes A \to A$ such that the two compositions $m\circ (1\otimes m)$ and $m\circ (m\otimes 1)$ agree as maps $A \otimes A \otimes A \to A$. (Actually, I always want to work with unital algebras, but I will surpress that part of the definition.) Any algebra $A = (A,m)$ has an opposite algebra $A^{\mathrm{op}} = (A,m\circ \text{flip})$, where $\text{flip}: A\otimes A \to A\otimes A$ takes $a\otimes b$ to $b\otimes a$. The 2-category $\mathrm{Alg}$ has as its objects the algebras. To call it a 2-category, I need to tell you a category $\hom(A,B)$ for any algebras $A$ and $B$. I declare that $\hom(A,B) = (A\otimes B^{\mathrm{op}})\text{-Mod}$.

What is this, you say? Well, one can naturally tensor algebras: given algebras $(A,m_A)$ and $(B,m_B)$, one declares that $(A,m_A)\otimes (B,m_B) = \bigl(A\otimes B, (m_A\otimes m_B) \circ (1\otimes \mathrm{flip} \otimes 1)\bigr)$. Given any algebra $A = (A,m)$, a left $A$-module is a vector space $V$ and a map $m : A\otimes V \to V$, such that $m\circ (1\otimes m) = m\circ (m\otimes 1)$ as maps $A\otimes A \otimes V \to V$. There is an obvious notion of homomorphism of left $A$-modules, and $A\text{-Mod}$ is the category of all left $A$-modules.

So, you say, we have a class of objects, and between any two objects $A$ and $B$ a category declared to be $\hom(A,B)$. That's not enough to define a 2-category! How do you compose 1-morphisms? This is a fair question, and I have a fair answer. Given $V \in \hom(A,B)$ and $W \in \hom(B,C)$, their composition $V\circ W = V\underset B \otimes W$ is defined as the quotient of $V \otimes W$ by the subspace spanned by elements of the form $m(v,b) \otimes w - v\otimes m(b,w)$ for $v\in V$, $w\in W$, and $b\in B$. Ah, you say, this is a good definition, and checking the rest is a reasonable exercise.

Looking over the construction, we see that we started with a symmetric monoidal 1-category $\text{Vect}$, and produced a symmetric monoidal 2-category $\text{Alg}$. It is very reasonable to try to repeat the trick. Notions like "associative algebra" and "module" make sense in any monoidal category. By definition, a sesquialgebra is an associative (and unital) algebra object in $\text{Alg}$. (The name "sesquialgebra" is from Tang, Weinstein, and Zhu, 2005. I believe the same definition was also found, independently and around the same time, by Douglas and Henriques, under the name "2-ring". It is quite likely that it has appeared many other times as well.) To unpack the definition a bit, a sesquialgebra is an algebra $A$, and a left $(A\otimes A \otimes A^{\mathrm{op}})$-module $M$, and also a module isomorphism $\alpha : M \underset{A \otimes A}\otimes (1 \otimes M) \cong M \underset{A \otimes A}\otimes (M\otimes 1)$, where here $1$ denotes $A$ as an $(A\otimes A^{\mathrm{op}})$-module and I'll let you work out which actions are tensored with which. The isomorphism $\alpha$ must satisfy a "pentagon" equation that you may look up. A left sesquimodule for a sesquialgebra $(A,M)$ is an algebra $V$ and a $(A\otimes V \otimes V^{\mathrm{op}})$-module $M$ and an isomorphism $\alpha$ satisfying the appropriate pentagon. Given a sesquialgebra $A$, there is a coopposite sesquialgebra $A^{\mathrm{coop}}$ which equals $A$ as an algebra, but with the actions of the module $M$ permuted. Examples of sesquialgebras include commutative algebras, Hopf algebras, and $A\otimes A^{\mathrm{op}}$ for any algebra $A$.

I have been told that the sesquialgebras are the objects of a 2-category $\text{SesquiAlg}$, in which $\hom(A,B)$ is the 2-category of $(A \otimes B^{\mathrm{coop}})$-sesquimodules. It is an easy enough exercise to write out the definition of this 2-category: objects are algebras with various modules, 1-morphisms are bimodules with various isomorphisms, etc.

But it is not at all clear to me how to actually compute the composition in $\text{SesquiAlg}$. I basically believe enough abstract nonsense to see that a composition should exist, but I'd like an explicit presentation akin to the one I gave above when you complained that I hadn't sufficiently defined $\text{Alg}$.

Question: Let $V$ be a left $(A\otimes B^{\mathrm{coop}})$-sesquimodule and $W$ a left $(B\otimes C^{\mathrm{coop}})$-sesquimodule. What is an explicit presentation for their composition $V \underset B \otimes W$? Among other things, all of $A,B,V,W$ are associative algebras — the bulk of my question asks for an explicit presentation of the associative algebra $V \underset B \otimes W$. Of course, as an associative algebra $V \underset B \otimes W$ defined only up to Morita equivalence, but I don't mind that — I just want a presentation for a representative of its Morita class.

It often happens that taking quotients is easier in higher-category land than in lower-category land ("the low country"?), because you often can present the quotient simply by adjoining some extra morphisms. Does something like that happen here?

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I think he means "associative (and unital) algebra object in $\text{Alg}$," although he hasn't picked out a tensor product on $\text{Alg}$ yet. – Qiaochu Yuan Jul 2 '13 at 5:11
@MarianoSuárez-Alvarez: Sorry, Alg. It's fixed now. – Theo Johnson-Freyd Jul 2 '13 at 5:20
@QiaochuYuan: I did define the tensor product of objects in Alg, and I leave it as an exercise to the "you" who speaks in italics in my question is extend this to 1- and 2-morphisms. The extension is uniquely determined by declaring that the tensor product of rank-one free modules is the rank-one free module in the obvious way, and that $\otimes$ should be cocontinuous on hom categories. – Theo Johnson-Freyd Jul 2 '13 at 5:22