Is the box product of morphisms associative?

Question

Suppose $(C,\otimes)$ is a symmetric monoidal finitely-cocomplete category such that $\otimes$ preserves colimits. Given two morphisms $a:A_1\to A_2$ and $b:B_1\to B_2$, define $a\Box b$ to be the induced "lower right corner map" $A_1\otimes B_2\coprod_{A_1\otimes B_1} A_2\otimes B_1\to A_2\otimes B_2$.

Is $\Box$ associative? Does it give a monoidal product on $Arr(C)$?

I'm having trouble figuring out what the diagram for this should look like if it's true (although a quick "arithmetic check" shows that it should be).

If it's not associative for some stupid reason that I've overlooked in the general case presented here, is it true in the case that $C$ is a presheaf category and $\otimes$ is the cartesian product?

Answer 1

Yes, that gives a monoidal product (but don't forget the monoidal unit! it's $0 \to I$ where $I$ is the monoidal unit of $\mathcal{C}$).

To simplify the description of the associativity, imagine that $\mathcal{C}$ is strict monoidal to begin with. Draw the entire cube of possible paths that lead from $A_1 \otimes B_1 \otimes C_1$ to $A_2 \otimes B_2 \otimes C_2$, except leave off the three edges that terminate at the corner $A_2 \otimes B_2 \otimes C_2$. The colimit of this diagram is what you want to consider. By taking pushouts one at a time (and taking advantage of preservation of colimits by tensoring on either side), you should be able to see that one way of deriving the colimit is by pushing out $(AB) \otimes C_1 \to A_2 \otimes B_2 \otimes C_1$ along $(AB) \otimes C_1 \to (AB) \otimes C_2$, and another way is by pushing out $A_1 \otimes (BC) \to A_2 \otimes (BC)$ along $A_1 \otimes (BC) \to A_1 \otimes B_2 \otimes C_2$.

I hope these hints convey the idea, because actually drawing the diagrams here would be a somewhat time-consuming job. :-)

Answer 2

I accepted Todd's answer because it was very helpful, but I'm going to write this answer because I figured out a very useful heuristic way to think about this (although I think that actually proving the equivalence wouldn't be very tough (or worthwhile)).

There is a very nice geometric description that makes the combinatorics of the problem much much clearer: Imagine that all of the maps involved are just the inclusion $0\to \mathbf{R}$. Then the box product of $n$ of these maps is the inclusion $X\hookrightarrow \mathbf{R}^n$, where $X$ is the union of all codimension 1 hyperplanes spanned by the axes. We can associate a commutative diagram with this description, which is generated by the different ways to glue over different maps, and this exactly represents the general problem. The reason why embedding in $\mathbf{R}^n$ in this way is so effective is that any two of the relevant objects are glued along their intersection. The embedding encodes the explicit combinatorial information so you don't need to worry about it.

In particular, without this picture (or a similar one) in mind, it will very quickly get extremely difficult to deal with box products of anything more than a few maps, since there are already tons of ways to glue together just three planes passing through the origin if you do it piece-by-piece. This is because using "pushout notation" as above, we not only have to keep track of the pieces we're gluing together, but also what we're gluing them along. If you don't understand the geometric picture, you're essentially proving it by brute force by drawing every single possible pushout path (see Todd's answer; indeed if there were more than three maps, the commutative diagram would be four dimensional) to the required map.

I hope people learn from my mistakes!