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?



Is the box product of morphisms associative?
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. :-) 


