Kai Bruennler asked: Is there a binary product in the category of sets and functions that is "strictly associative", i.e. A x (B x C) = (A x B) x C and the associativity isomorphisms are equal to the identity? I gave a construction using the axiom of choice. Well, we can do better. Fix on an ordered-pair construction, say Weiner's, and denote its values as <x,y>. Let l be the left-coordinate function, r the right-coordinate function. Fix on a sequence of finite ordinals, say von Neumann's. Define, inductively, an n-TUPLE as something of the form <x,y> where y is a (n-1)-tuple if n>1 else any old set if n=1. (There are no 0-tuples.) Note that the same set can be an n-tuple for any number of values for n. But if x is an n-tuple and we are given n then each of it's n coordinates is well-defined: x_1 = lx x_2 = l(rx) ... x_{n-1} = l(r(r(...r(rx)...))) x_n = r(r(r(...r(rx)...))) where there are i-1 applications of r used for x_i, one application of l used for x_i if i<n and no application of l used for x_n. A trick of this construction of binary products is that we define products twice. Given n sets a_1, a_2,...,a_n define their n-fold PRE-PRODUCT as the set of pairs of the form <x,n> where x is an n-tuple such that x_i \in a_i each relevant i. If two sets x and y arise as pre-products, that is, if there are positive ordinals m and n and sets a_1, a_2,..., a_m, b_1, b_2, ...,b_n such that x is the pre-product of the a's and y is the product of the b's, then we define their PRODUCT as the (m+n)-fold pre-product of a_1, a_2,..., a_m, b_1, b_2,...,b_n. Define F(x) = x if x is a pre-product else { <y,1> | y \in x}. Define the product of arbitrary x and y as the product of F(x) and F(y).