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 XYX \coprod Y is empty. In this case one writes XYX+YX \coprod Y \coloneqq X + Y for the coproduct and speaks of the disjoint union of XX with YY.


In a category

A 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\coprod_i a_i is disjoint if each coprojection a i ia ia_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.


  • 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 {01}\left\{ 0 \to 1 \right\} has coproducts but is 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.


For instance page 34 in section A1.4.4 in

Last revised on October 5, 2020 at 15:48:13. See the history of this page for a list of all contributions to it.