nLab
distributive category

Contents

Context

Monoidal categories

monoidal categories

With symmetry

With duals for objects

With duals for morphisms

With traces

Closed structure

Special sorts of products

Semisimplicity

Morphisms

Internal monoids

Examples

Theorems

In higher category theory

Category theory

Contents

Definition

Definition

A category CC with finite products ()×()(-)\times(-) and coproducts ()+()(-) + (-) is called (finitary) distributive if for any X,Y,ZCX,Y,Z\in C the canonical distributivity morphism

X×Y+X×ZX×(Y+Z) X\times Y + X\times Z \longrightarrow X\times (Y+Z)

is an isomorphism. The canonical morphism is the unique morphism such that X×YX×(Y+Z)X\times Y \to X\times (Y+Z) is X×iX\times i, where i:YY+Zi\colon Y\to Y +Z is the coproduct injection, and dually for X×ZX×(Y+Z)X\times Z \to X\times (Y+Z).

Remark

This notion is part of a hierarchy of distributivity for monoidal structures, and generalizes to distributive monoidal categories and rig categories. A linearly distributive category is not distributive in this sense.

This axiom on binary coproducts easily implies the analogous nn-ary result for n>2n\gt 2. In fact it also implies the analogous 0-ary statement that the projection

X×00 X\times 0 \to 0

is an isomorphism for any XX (see Proposition below). Moreover, for a category with finite products and coproducts to be distributive, it actually suffices for there to be any natural family of isomorphisms X×Y+X×ZX×(Y+Z)X\times Y + X\times Z \cong X\times (Y+Z), not necessarily the canonical ones; see the paper of Lack referenced below.

A category CC with finite products and all small coproducts is infinitary distributive if the statement applies to all small coproducts. One can also consider κ\kappa-distributivity for a cardinal number κ\kappa, meaning the statement applies to coproducts of cardinality <κ\lt\kappa.

Any extensive category is distributive, but the converse is not true.

Properties

Proposition

In a category with products and coproducts, if products distribute over binary coproducts, then coproduct coprojections are monic.

Proof

Let i B:BB+Ci_B: B \to B + C be a coproduct coprojection, and suppose given maps f,g:ABf, g: A \to B such that i Bf=i Bgi_B f = i_B g. We observe that the coprojection

i:A×BA×B+A×Ci: A \times B \to A \times B + A \times C

is monic because it has a retraction (1 A×B,ϕ):A×B+A×CA×B(1_{A \times B}, \phi): A \times B + A \times C \to A \times B. (All we need here is the existence of a map ϕ:A×CA×B\phi: A \times C \to A \times B, for example the composite A×Cπ AA1 A,fA×BA \times C \stackrel{\pi_A}{\to} A \stackrel{\langle 1_A, f \rangle}{\to} A \times B.)

The composite of the coprojection ii with the canonical isomorphism A×B+A×CA×(B+C)A \times B + A \times C \cong A \times (B + C), namely 1 A×i B:A×BA×(B+C)1_A \times i_B: A \times B \to A \times (B + C), is therefore also monic. Given that 1 A,i Bf=1 A,i Bg:AA×(B+C)\langle 1_A, i_B f \rangle = \langle 1_A, i_B g \rangle: A \to A \times (B + C), we conclude

(1 A×i B)1 A,f=1 A,i Bf=1 A,i Bg=(1 A×i B)1,g,(1_A \times i_B)\langle 1_A, f \rangle = \langle 1_A, i_B f \rangle = \langle 1_A, i_B g \rangle = (1_A \times i_B)\langle 1, g \rangle,

whence 1 A,f=1 A,g:AA×B\langle 1_A, f\rangle = \langle 1_A, g\rangle: A \to A \times B since 1 A×i B1_A \times i_B is monic. It follows that f=gf = g, as was to be shown.

Proposition

If products distribute over binary coproducts, then products distribute over nullary coproducts (i.e., the projection X×00X \times 0 \to 0 is an isomorphism for all objects XX).

Proof

Clearly hom(X×0,Y)\hom(X \times 0, Y) is inhabited by X×00YX \times 0 \to 0 \to Y for any object YY. On the other hand, since the two coprojections 00+00 \to 0 + 0 coincide, the same holds for the two coprojections X×0(X×0)+(X×0)X \times 0 \to (X \times 0) + (X \times 0), by applying the distributivity isomorphism X×(0+0)(X×0)+(X×0)X \times (0 + 0) \cong (X \times 0) + (X \times 0). This is enough to show that any two maps X×0YX \times 0 \to Y coincide.

Proposition

In a distributive category, the initial object is strict.

Proof

Given an arrow f:A0f: A \to 0, we have that π A:A×0A\pi_A: A \times 0 \to A is a retraction of 1,f:AA×0\langle 1, f \rangle: A \to A \times 0, so that AA is a retract of A×00A \times 0 \cong 0. But retracts of initial objects are initial.

References

Last revised on May 3, 2016 at 15:27:10. See the history of this page for a list of all contributions to it.