nLab disjoint coproduct

Idea

The notion of disjoint coproduct is a generalization to arbitrary categories of that of disjoint union of sets.

One says that a coproduct $X + Y$ of two objects $X, Y$ in a category $\mathcal{C}$ is disjoint if the intersection of $X$ with $Y$ in $X + Y$ is empty. In this case one often writes $X \coprod Y \coloneqq X + Y$ for the coproduct, particularly if the coproduct is stable under pullbacks, and there one speaks of the disjoint union of $X$ with $Y$.

Definition

In a category

A coproduct $a+b$ in a category is disjoint if

1. the coprojections $a\to a+b$ and $b\to a+b$ are monic, and

2. their intersection is an initial object.

Equivalently, this means we have pullback squares

$$\array{ a & \to & a &&& b & \to & b &&& 0 & \to & b\\ \downarrow && \downarrow &&& \downarrow && \downarrow &&& \downarrow && \downarrow \\ a & \to & a+b &&& b & \to & a+b &&& a & \to & a+b}$$

An arbitrary coproduct $\sum_i a_i$ is disjoint if each coprojection $a_i\to \sum_i a_i$ is monic and the intersection of any two is initial. Note that every 0-ary coproduct (that, is initial object) is disjoint.

Examples

• In the category $Set$ of sets and functions, the coproduct is given by disjoint union and is, unsurprisingly, disjoint. In the category $Pfn$ of sets and partial functions the coproduct is equally given by disjoint union and total injections and is disjoint as well.

• Since having all finitary disjoint coproducts is half of the condition for a category to be extensive, extensive categories provide examples for categories with disjoint finite coproducts. In the preceding discussion $Set$ instantiates this case whereas $Pfn$ does not: since the initial object $\emptyset$ in $Pfn$ is not strict, the latter category is not extensive.

• In the category $Vect$ of (real) vector spaces coproducts are given by direct sum and are disjoint but not stable under pullback: pulling back the colimit diagram $\mathbb{R}\to\mathbb{R}\oplus\mathbb{R}\leftarrow\mathbb{R}$ along the diagonal morphism $\Delta:\mathbb{R}\to\mathbb{R}\oplus\mathbb{R}\; ,\; x\mapsto x\oplus x$ yields $0\to\mathbb{R}\leftarrow 0$ which is not a colimit diagram. Whence $Vect$ is not extensive.

• Non-example: the interval category $\left\{ 0 \to 1 \right\}$ has coproducts but they are not all disjoint: $1+1=1$. There are plenty more examples of posets that have non-disjoint coproducts besides this one. In a Boolean algebra, two elements $a$ and $b$ are disjoint in the sense that $a\wedge b=0$ if and only if $a\vee b$ is their disjoint coproduct.

Properties

Characterization of sheaf toposes

Having all small disjoint coproducts is one of the conditions in Giraud's theorem characterizing sheaf toposes.

In coherent categories

Proposition

Let $\mathcal{C}$ be a coherent category. If $X, Y \hookrightarrow Z$ are two subobjects of some object $Z \in \mathcal{C}$ and are disjoint, in that their intersection in $Z$ is empty, $X \cap Y \simeq\emptyset$, then their union $X \cup Y$ is their (disjoint) coproduct.

This apears as (Johnstone, cor. A1.4.4).

Definition

A coherent category in which all coproducts are disjoint is also called a positive coherent category.

(Johnstone, p. 34)

Example

Every extensive category is in particular positive, by definition.

In a positive coherent category, every morphism into a coproduct factors through the coproduct coprojections:

Proposition

Let $\mathcal{C}$ be a postive coherent category, def. , and let $f \colon A \to X \coprod Y$ be a morphism. Then the two subobjects $f^*(X) \hookrightarrow A$ and $f^*(Y) \hookrightarrow Y$ of $A$, being the pullbacks in

$$\array{ f^* (X) &\to& X \\ \downarrow && \downarrow^{\mathrlap{i_X}} \\ A &\stackrel{f}{\to}& X \coprod Y } \;\;\;\; \array{ f^* (Y) &\to& Y \\ \downarrow && \downarrow^{\mathrlap{i_Y}} \\ A &\stackrel{f}{\to}& X \coprod Y }$$

are disjoint in $A$ and $A$ is their disjoint coproduct

$$A \simeq f^*(X) \coprod f^*(Y) \,.$$

This appears in (Johnstone, p. 34).

Remark

This means that if $A \in \mathcal{C}$ itself is indecomposable in that it is not a coproduct of two objects in a non-trivial way, for instance if $\mathcal{C}$ is an extensive category and $A \in \mathcal{C}$ is a connected object, then every morphism $A \to X \coprod Y$ into a disjoint coproduct factors through one of the two canonical inclusions.

Related concepts

• disjoint subset

• disjunctive logic

References

• Aurelio Carboni, Stephen Lack, R. F. C. Walters, around Def. 2.5 in: Introduction to extensive and distributive categories, JPAA 84 (1993) pp. 145-158 (doi:10.1016/0022-4049(93)90035-R)

• Peter Johnstone, Sec. A1.4.4, p. 34 in: Sketches of an Elephant