Jamie Vicary wrote:
Of course, in the light of http://www.arxiv.org/abs/math.CT/0604542 , perhaps we only need require that our dagger-category has products and equalizers in order for it to have 'finite bilimits'! In remark 2.6 of [2] cited in your email below, you write that if a dagger-category has products then it must of course have coproducts, but it need not have biproducts. Presumably, math.CT/0604542 proves you wrong here?
You are referring to the paper "Finite Products are Biproducts in a Compact Closed Category" by Robin Houston. It does not prove me wrong. Robin's construction only applies to compact closed categories. In general, a dagger category doesn't need to be compact closed. Actually, there is a counterexample to support my remark 2.6. It is due to Robin Houston and myself (any typos or mistakes are mine). (1) There exists a category C with finite products and coproducts, with a zero object, and such that for all A,B, A+B is isomorphic to AxB (not naturally), but for some A,B, the canonical map f:A+B -> AxB is not an isomorphism. Proof: Let C be the category of sets of cardinality 0 or aleph_0, with partial functions as the morphisms. Then the empty set is initial and terminal. We have AxB = A \union (A*B) \union B, where "x" denotes categorical product, and "*" denotes cartesian product of sets. Further A+B = A \union B. By inspecting cardinalities, we find that AxB and A+B have equal cardinality, for all A, B, and hence they are (not naturally) isomorphic. However, the canonical map f:A+B -> AxB satisfying p_i.f.q_j = \delta_{ij} maps everything to the first and third components of AxB = A \union (A*B) \union B, hence is not onto when A,B are non-empty. (2) Corollary: C does not have biproducts. (3) Corollary: There exists a category C with finite products and coproducts, and such that A+B = AxB for all A,B, but which does not have biproducts. Proof: choose a skeleton of the category in (1). (4) There exists a dagger category with finite products, but which does not have biproducts. Proof: Take C as in (3), and consider D = C x C^op. Then D has products and coproducts as inherited componentwise from C and C^op. Also, it satisfies X+Y = XxY. Further, D has no biproducts, or else the forgetful functor to C would preserve them. Now consider D', the full subcategory of D consisting of objects of the form (A,A). This has products and coproducts, and they are not biproducts. Further, D' is a dagger category with (f,g)^{\dag} = (g,f). Another related remark is that even *if* a dagger category has biproducts, then they need not be dagger-biproducts. Here is a counterexample. Consider the category of matrices with rational entries. Objects are arities, and composition is standard matrix multiplication, but define the following non-standard dagger: if A is an mxn-matrix, then let A^\dag = A^{transpose} * 3^(n-m). This is indeed an involutive, identity-on-objects functor. As a category, it is equivalent to finite-dimensional Q-vector spaces, so it has biproducts, and it is also compact closed. However, there are no isometries e : Q -> Q^2 (and hence, no dagger-biproducts). If such an isometry existed, say with matrix (a, b)^{transpose}, then we would have e^\dag.e = (a^2 + b^2)/3 = 1. However, the equation a^2 + b^2 = 3 has no solution in the rational numbers. (In Z/9Z, any sum of two squares that is divisible by 3 is also divisible by 9; therefore the same holds in the integers. The claim about the rational numbers follows by taking a sufficiently large square common denominator). I do not know whether Robin Houston's construction, when applied to a dagger compact closed category, yields dagger-biproducts. Note that the previous counterexample is not dagger-compact closed (dagger does not preserve tensor). It therefore doesn't answer this last question. Best, -- Peter