OOPS! In my previous posting i've made some mixed-up and talked mainly about notation for diagrams while the actual question is whether product is commutative (symmetric) operation in Sets. Below is a sketch of possible solution. First of all, product is a diagram operation and explicit presentation of their syntax would be useful; so, let me present a bit of technicalities. Let G be some base category (think of the category of graphs). The arity of a diagram operation F (think of products, push-outs...) is a G-morphism s: S^in --> S with S the shape of the operation and S^in the input shape. For example, for binary products, S is the graph [x] <--p-- [u] --q--> [y] where [ ]'s denote nodes and arrows' names are written right on the arrows, S^in is the two-node discrete graph [x] [y], and morphism s is its embedding into S. A G-object C (think of a category) is an F-algebra if for any diagram d: S^in --> C there is a unique diagram F^C(d): S-->C such that s;F^C(d) = d (input data are preserved); below i'll write just F for F^C. Well, categorically we normally have many isomorphic F(d)'s, let's skip this for a while. And as for products of sets (ie, F is Prod, C is Sets), by using names p and q that are already given in the shape, we really have a canonical choice. Namely, for any d: [x] [y] --> Sets with x.d=A, y.d=B, we define u.Prod(d) = { t: {p,q} --> A \cup B | p.t \in A, q.t \in B } with evident p.Prod(d) and q.Prod(d). F is called commutative on algebra C if any isomorphism i: S^in --> S^in has an isomorphic extension i+: S --> S with s;i+ = i;s s.t. for any d: S^in --> C, F(i;d) = i+ ; F(d) Now it is seen that Prod is commutative in Sets because any isomorphism i: S^in --> S^in has an evident required extension i+, i+(u) = u and i+ acts on the arrows p,q, . exactly like i acts on their targets x,y,.. The illusion of non-commutativity of products in Sets arises because of disregarding the diagrammatic nature of the operation: isomorphism i on the nodes in the shape acts on d but the corresponding i+ acting on the arrows is not taken into consideration. A peculiarity of what was described above is that names of items in the shape are used in the definition of the operation. So, for example, the following two shapes [x] <-- weight -- [u] -- height --> [y] and [x] <-- red -- [u] - blue -->[y] though isomorphic yet different and determine the two different product operations producing isomorphic yet different results from the same input. Thus, what we usually call the product operation is something like operation schema parameterized by pairs of names. Then expressions like Prod[weight:A, height:B] denote a concrete instance of the shape schema and, simultaneously, a diagram d in Sets with weight.codom.d = A and height.codom.d = B. Now back to AxB vs. BxA. The writing AxB is normally unambiguous, it means [1st:A] x [2nd:B] and encodes (i) the standard shape instance [x] <-- 1st -- [u] -- 2nd --> [y] and (ii) the diagram d: [x] [y] --> Sets, x.d = A, y.d=B. In contrast, the writing BxA is really ambiguous: it may mean either (a) [1st:B] x [2nd:A] that encodes the same standard shape instance [x] <-- 1st -- [u] -- 2nd --> [y] but now the diagram is x.d = B, y.d=A; or it may mean (b) [2nd:A] x [1st:B] that encodes the non-standard shape instance [x] <-- 2nd-- [u] -- 1st --> [y] and again the diagram x.d = B, y.d=A. It seems that thinking naïvely-set-theoretically people tend to see in writing BxA the case (a) and then, of course, BxA is not equal to AxB. And thinking categorically, people tend to see in writing BxA the case (b) and then BxA is the same as AxB. So, the problem 'ordered vs. unordered products ' is a problem of poor notation for product operation like BxA while the operation itself is a *commutative* diagram operation as defined above. Actually it was already said in early postings but now it became more transparent. Finally, some speculation. It seems that parameterization of (shapes of ) operations by names, and use of these names in operation's definitions, is applicable to other standard categorical operations and can be considered in a general setting. Probably, it could be made precise by some kind of internalization ...? Zinovy Diskin