Let $\Delta$ be the simplex category. For $m \leq c \leq n$, let $[m] \to [c] \to [n]$ be the composition of two injective morphisms in $\Delta$.

We now define a category $\Box$ with same objects as $\Delta$, but we think of them as cubes instead of simplices. Combinatorially, the $m$-facets of the $n$-cube $[n]$ should be exactly compositions of the form $[m] \to [c] \to [n]$ as above. Thus, I would like to define

$$\Box_m([m],[n]) := \coprod_{c=m}^n ( \Delta_m([c],[n]) \times \Delta_m([m],[c]) ), $$ where the index $m$ indicates that I am considering only monomorphisms. I'm not a category theory expert, but this looks a lot like a coend over the product of hom-functors: $$F(\cdot,\cdot):=( \Delta_m(\cdot,[n]) \times \Delta_m([m],\cdot) ) \qquad \Rightarrow \qquad \Box_m([m],[n]) := \int^{[c] \in \Delta} F([c],[c]).$$ My question is basically, if this observation - i.e. redefining the morphism set in a category by taking a coend of the previous form - is helpful in the following sense:

1. It would be nice if the above definition would automatically give me a composition law by some categorical nonsense
2. It would also be nice if the above definition would guide the way for defining general morphisms (simply omit '$m$').

Both points can of course be done directly without a lot of effort. Still it would be a nice confirmation if the construction turns out to be 'natural' in the categorical sense.

edit:

If I understand it correctly, by the co-Yoneda lemma the coend above is just $\Delta_m([m],[n])$... thinking about it, taking the coend quotient is exactly the opposite of what I want. This question can actually be closed as it doesnt make much sense, sorry..

edit:

The above coproduct is actually not completely correct. I think of a map in $\Delta_m([m],[n])$ as a composition

$$ [m] \to [m] \oplus [n - m] \xrightarrow{\sigma_{m,n-m}} [n], $$ where the first map is the canonical inclusion and $\sigma_{m,n-m} \in S_n$ is a shuffle. Note, that there is a choice for the inclusion involved here. For simplices it didn't matter, but for cubes, it does: in a similar spirit as above a map in $\Box_m([m],[n])$ is given by a composition

$$[m] \to [c-m] \oplus [m] \xrightarrow{\sigma_{c-m,m}} [c] \to [c] \oplus [n-c] \xrightarrow{\sigma_{c,n-c}} [n] $$

or simply $$[m] \to [c-m] \oplus [m] \oplus [n-c] \xrightarrow{\sigma_{c-m,m,n-c}}[n], $$ where $\sigma_{c-m,m,n-c}$ is a 'trinomial' shuffle (not sure if this is the right term, but you get the idea). Thus, the definition of morphisms in $\Box$ in terms of morphisms in $\Delta$ should actually be: $$ \Box_m([m],[n]) := \coprod_{c=m}^n ( \Delta_m([c],[n]) \times \Delta_m([c-m],[c]) ).$$ So given shuffles $$ f=\sigma_{c-l,l,m-c,} \in \Box_m([l],[m]), \qquad g=\sigma_{d-m,m,n-d} \in \Box_m([m],[n]),$$ there is a unique shuffle $$g\circ f =\sigma_{c+d-(l+m),l,n+m-(c+d)} \in \Box_m([l],[n]),$$ such that (not yet sure).