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 $Y$.
A coproduct $a+b$ in a category is disjoint if
the coprojections $a\to a+b$ and $b\to a+b$ are monic, and
their intersection is an initial object.
Equivalently, this means we have pullback squares
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.
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.
Having all small disjoint coproducts is one of the conditions in Giraud's theorem characterizing sheaf toposes.
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).
A coherent category in which all coproducts are disjoint is also called a positive coherent category.
Every extensive category is in particular positive, by definition.
In a positive coherent category, every morphism into a coproduct factors through the coproduct coprojections:
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
are disjoint in $A$ and $A$ is their disjoint coproduct
This appears in (Johnstone, p. 34).
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.
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
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