Re: Equalisers and coequalisers in categories with a \dag-involution

[selinger@mathstat.dal.ca (Peter Selinger)]

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