# nLab disjoint coproduct

Contents

### Context

#### Limits and colimits

limits and colimits

# Contents

## 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 $X \coprod Y$ is empty. In this case one writes $X \coprod Y \coloneqq X + Y$ for the coproduct and 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 $\coprod_i a_i$ is disjoint if each coprojection $a_i\to \coprod_i a_i$ 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 of sets, all coproducts are disjoint.
• A category having all finitary disjoint coproducts is half of the condition for a category to be extensive. For further concrete examples of categories where all coproducts are disjoint, see extensive category.
• Non-example: the interval category $\left\{ 0 \to 1 \right\}$ has coproducts but is 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.

###### 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.

For instance page 34 in section A1.4.4 in