An n-category is an ordered list of n categories, any two of which (in the given order) form a 2-category. An n-functor is n functors between two n-categories. The categorical product axb of two m-categories a and b is an m-category. I want a product a@b yielding the "obvious" 2m-category (or (m+n)-category when a is an m-category and b an n-category). I've been trying to figure out from papers of Street, Aitchison, Johnson, etc., how this might be accomplished, but thus far I haven't got the picture, their reassuring hints that it's doable notwithstanding. What am I missing? What's the trick? Where should I look? I presume I should be aiming for a tensor product rather than a categorical product. Is there some routine construction of homological algebra or tarot card theory that does the trick? In more detail: if every story has a beginning, a middle, and an end, then the square of a story has 9 parts, the "biggest" of which is a pair (m,m) of middles. For the categorical square axb this is the arrow from (b,b) to (e,e), a morphism or 1-cell. In a@b, (m,m) should be a solid square or 2-cell. I'm assuming a@b will be oriented. Indeed I'm expecting a@b will be 1-isomorphic to but "2-antimorphic" to b@a (the two functors from a@b to b@a will be covariant and contravariant respectively). I also don't mind if this square ends up with more than 9 things, in fact I'm vaguely expecting 11 (we need the two composites at the start and end of the square) but I'll settle for even more if that's what the construction takes. If the answer is 42 I'll go into some other line of work. Writing ab and a*b for the respective compositions of a 2-category (namely horizontal and vertical composition), it would be nice if a@(bc) = ((a@b)c) * (b(a@c)), as suggested by the following picture. b c -----------> ---------> | ---------> | | | b c | b | | __ | a| @ ------>------> = | __ | |a //| |a | a| //| a| | // | | | // | V c V V V V ---------> ---------> -----------> b c Here |a| = 2+1 (i.e. a has 2 0-cells and 1 1-cell), |bc| = 3+3, and |a@(bc)| = 6+16+5 (6 0-cells, 16 1-cells, and 5 2-cells). (Counting all composites, but not counting degenerate cells, i.e. n-cells of dimension less than n, these having already been counted at the lower dimension. For a@(bc), don't just count cells in the picture on the right, which has 10 0-cells, paste first to get down to 6 0-cells.) And (bc)@a should equal (b(c@a)) * ((b@a)c). b c -----------> ---------> | ---------> | | b c | | b | | // | ------>------> @ a| = | | |a |// |a | a| // a| | -- | | | |// | V c V V V -- V ---------> ---------> -----------> b c -Vaughan Pratt