nLab disjoint coproduct




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

One says that a coproduct X+YX + Y of two objects X,YX, Y in a category 𝒞\mathcal{C} is disjoint if the intersection of XX with YY in X+YX + Y is empty. In this case one often writes XYX+YX \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 XX with YY.


In a category

A binary coproduct a+ba+b in a category is disjoint if

  1. the coprojections aa+ba\to a+b and ba+bb\to a+b are monic, and

  2. their intersection is an initial object.

Equivalently, this means we have pullback squares

a a b b 0 b a a+b b a+b a a+b \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 ia i\sum_i a_i is disjoint if each coprojection a i ka ka_i\to \sum_k a_k is monic and the intersection of any two distinct ones is initial. Note that every 0-ary coproduct (that, is initial object) is disjoint.

A more constructive way to phrase disjointness of an arbitrary coproduct is that the pullback of any two coprojections a i ka ka_i\to \sum_k a_k and a j ka ka_j\to \sum_k a_k is the coproduct i=ja i\sum_{i=j} a_i, where i=ji=j denotes the subsingleton corresponding to the proposition i=ji=j, a.k.a. {*i=j}\{ \ast \mid i=j \}. (Since a i=a ja_i=a_j as soon as this indexing set is inhabited, this coproduct could equally be written i=ja j\sum_{i=j} a_j.)


  • In the category SetSet of sets and functions, the coproduct is given by disjoint union and is, unsurprisingly, disjoint. In the category PfnPfn 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 SetSet instantiates this case whereas PfnPfn does not: since the initial object \emptyset in PfnPfn is not strict, the latter category is not extensive.

  • In the category VectVect 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 Δ:,xxx\Delta:\mathbb{R}\to\mathbb{R}\oplus\mathbb{R}\; ,\; x\mapsto x\oplus x yields 000\to\mathbb{R}\leftarrow 0 which is not a colimit diagram. Whence VectVect is not extensive.

  • Non-example: the interval category {01}\left\{ 0 \to 1 \right\} has coproducts but they are not all disjoint: 1+1=11+1=1. There are plenty more examples of posets that have non-disjoint coproducts besides this one. In a Boolean algebra, two elements aa and bb are disjoint in the sense that ab=0a\wedge b=0 if and only if aba\vee b is their disjoint coproduct.


Characterization of sheaf toposes

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

In coherent categories


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

(Johnstone, p. 34)


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

f *(X) X i X A f XYf *(Y) Y i Y A f XY \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 AA and AA is their disjoint coproduct

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

This appears in (Johnstone, p. 34).


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


Last revised on July 27, 2023 at 07:28:58. See the history of this page for a list of all contributions to it.