nLab extensive category

Idea

A category is extensive if it has coproducts that interact well with a certain class of pullbacks. Variations (some only terminological) include lextensive, disjunctive, and positive categories. All of these come in finitary and infinitary versions (and, more generally, $\kappa$-ary versions for any arity class $\kappa$).

Definitions

Extensive categories

Definition

A finitely extensive category (or finitary extensive category) is a category $E$ with finite coproducts such that one, and hence all, of the following equivalent conditions holds:

1. For any pair of objects $a, b$ the coproduct functor on slice categories is an equivalence of categories:

   (1)$$\array{ E_{/a} \,\times\, E_{/b} &\xrightarrow{\;\;\;\simeq\;\;\;}& E_{/(a \sqcup b)} \\ \left( \array{ X \\ \downarrow \\ a } \right) ,\; \left( \array{ Y \\ \downarrow \\ b } \right) &\mapsto& \left( \array{ X \sqcup Y \\ \downarrow \\ a \sqcup b } \right) }$$

   (e.g. Carboni, Lack & Walters 1993, Def. 2.1).

2. Pullbacks of finite-coproduct injections along arbitrary morphisms exist and finite coproducts are disjoint and stable under pullback.

   (e.g CLW93, Prop. 2.6, Lem. 2.11)

3. Pullbacks of finite-coproduct injections (and thus all coproduct injections) along arbitrary morphisms exist, and in any commutative diagram

   (2)$$\array{ x & \to & z & \leftarrow & y \\ \downarrow & & \downarrow & & \downarrow \\ a & \to & a+b & \leftarrow & b }$$

   the two squares are pullbacks if and only if the top row is a coproduct diagram

   (e.g. CLW93, Prop. 2.2).

4. Finite coproducts are van Kampen colimits. By definition, this is just to say that one of the previous two conditions holds.

Definition

An infinitary extensive category is a category $E$ with all (small) coproducts such that the following analogous equivalent conditions hold:

1. Pullbacks of coproduct injections along arbitrary morphisms exist

   and small coproducts are disjoint and stable under pullback.

2. For any small-indexed set $(a_i)$ of objects, the coproduct functor of slice categories, generalizing (1), is an equivalence of categories:

   $$\underset{i \in I}{\prod} (E_{/a_i}) \overset{\sim}{\longrightarrow} E_{/(\coprod_i a_i)} \,.$$

3. Pullbacks of finite-coproduct injections (and thus all coproduct injections) along arbitrary morphisms exist, and for any family of commutative squares

   $$\array{ x_i & \to & z \\\downarrow &&\downarrow^f \\ a_i & \to & \coprod_i a_i }$$

   in which the bottom family of morphisms is the coproduct injections and the right-hand morphism is always the same, the top family are the injections of a coproduct diagram (hence $z = \coprod_i x_i$) if and only if all the squares are pullbacks.

4. All small coproducts are van Kampen colimits.

Remark

In between finitary and infinitary extensive , a $\kappa$-ary extensive category (for $\kappa$ a cardinal number or an arity class) is one with disjoint and stable coproducts of fewer than $\kappa$ objects. The unqualified term extensive category can refer to either the finitary or infinitary version, depending on the author; the more usual meaning is the finitary version.

Remark

Extensive categories are also called positive categories, especially if they are also coherent. Note that any disjoint coproduct in a coherent category is automatically pullback-stable. A positive coherent category which is also exact is called a pretopos. Infinitary pretoposes encapsulate all the exactness conditions of Giraud’s theorem characterizing Grothendieck toposes (the remaining condition is the existence of a small generating set).

Definition

If an extensive category also has finite limits, it is called lextensive or disjunctive.

(Note that the more usual default meaning of ‘disjunctive’, unlike the other terms, is the infinitary case.)

Remark

The alternative definitions of finitary disjunctive refer only to binary coproducts, but they obviously imply analogous statements for $n$-ary coproducts for all finite $n \ge 1$. Less obviously, they also imply the analogous statement for $0$-ary coproducts (that is, initial objects). In this case, the statement is that the initial object 0 is strict (any map $a\to 0$ is an isomorphism).

Furthermore, if binary coproducts are disjoint, then (at least assuming classical logic) any infinitary coproducts that exist are also disjoint, since

$$\bigsqcup_{a\in A} X_a \;\;\cong\;\; X_{a_0} \sqcup \bigsqcup_{a\neq a_0} X_a \;\;\cong\;\; X_{a_0} \sqcup X_{a_1} \sqcup \bigsqcup_{a\neq a_0,a_1} X_a$$

for any $a_0, a_1\in A$. Therefore, if a finitary-extensive category has infinitary pullback-stable coproducts, it is necessarily infinitary-extensive. In particular, a cocomplete locally cartesian closed category is finitary extensive if and only if it is infinitary extensive.

Remark

As a further special case of the preceding, since an elementary topos is finitary extensive, any cocomplete elementary topos is infinitary extensive. However, in this case, one of the arguments for finitary extensivity applies directly to the infinitary case and does not require classical logic; see toposes are extensive.

Remark

See familial regularity and exactness for a generalization of extensivity and its relationship to exactness.

Remark

Any extensive category with finite products is automatically a distributive category.

Superextensive sites

Any extensive category admits a Grothendieck topology whose covering families are (generated by) the families of inclusions into a coproduct (finite or small, as appropriate). We call this the extensive coverage or extensive topology. The codomain fibration of any extensive category is a stack for its extensive topology.

In general, we call a site superextensive if its underlying category is extensive, its covering families are generated by (finite or small) families, and its coverage includes the extensive one. See superextensive site for more details.

Pre-lextensive categories

Extensivity is an “exactness” condition, analogous to being a exact category or a pretopos (a pretopos being precisely a category that is exact and finitary-extensive). The corresponding “regularity” condition analogous to being a regular category or a coherent category is not well-known, but is not hard to write down.

Let us say (without making any assertion that this is good or permanent terminology) that a category is pre-lextensive if

1. it has a strict initial object $0$ (equivalently, its subobject preorders have pullback-stable bottom elements), and
2. whenever $A\rightarrowtail X$ and $B\rightarrowtail X$ are disjoint subobjects (i.e. $A\cap B=0$), they have a pullback-stable union (which is then automatically a disjoint and stable coproduct).

This is intended to complete the table of analogies:

some	all
regular category	exact category
coherent category	pretopos
pre-lextensive category	lextensive category

Regular/exact categories have quotients of (some) congruences. Exact categories have quotients of all congruences, while regular ones have quotients only of congruences that are kernel pairs. Another way to say that is that in a regular category, you can conclude that the quotient of some congruence exists if you can exhibit another object of which the quotient would be a subobject if it existed. Similarly, pre-/lextensive categories have disjoint unions. Lextensive categories have all disjoint unions (= coproducts), while in a pre-lextensive category you can conclude that a pair of objects $X,Y$ have a disjoint union if you can exhibit another object in which $X$ and $Y$ can be embedded disjointly. Finally, coherent categories/pretoposes have both quotients and disjoint unions, or equivalently quotients and not-necessarily-disjoint unions, with the same sort of relationship between the two.

Evidently a pre-lextensive category is lextensive as soon as any two objects can be embedded disjointly in a third. Pre-lextensive categories also suffice for the interpretation of disjunctive logic.

Being pre-lextensive is also sufficient to define the extensive topology and show that it is subcanonical, since it implies that whatever disjoint coproducts exist must be pullback-stable. The codomain fibration of a pre-lextensive category is not necessarily a stack for its extensive topology, but this condition is weaker than extensivity. It is true, however, that if $C$ is a pre-lextensive category whose codomain fibration is a stack for its extensive topology, and in which the disjoint coproduct $1+1$ exists, then $C$ is extensive, for arbitrary disjoint (binary) coproducts can then be obtained by descent along the covering family $(1\to 1+1, 1\to 1+1)$.

Examples

Example

Any free coproduct completion is extensive.

As one immediately checks condition (1), e.g. Carboni, Lack & Walters 1993, Prop. 2.4.

Example

An elementary topos (or, more generally, any pretopos) is finitary lextensive; a Grothendieck topos (or, more generally, any cocomplete elementary topos) is infinitary lextensive.

Example

A quasitopos with disjoint coproducts, or more generally a locally cartesian closed category with disjoint coproducts, is extensive. (Of course not all quasitoposes have disjoint coproducts, one example being a complete Heyting algebra.)

Example

The category Top of topological spaces is infinitary lextensive. The category Diff of smooth manifolds is infinitary extensive, though it does not have all pullbacks (only those involving a cospan of transversal maps).

Example

The category of schemes is infinitary lextensive. In more detail: the category of functors $CRing \to Set$ is infinitary lextensive (since finite limits and small coproducts are computed pointwise in $Set$), then sheaves with respect to the Zariski topology on $CRing^{op}$ form an infinitary lextensive category (since finite limits and small coproducts are reflected back from $[CRing, Set]$ by applying a left exact reflection to the inclusion of sheaves in presheaves). Finally, the category of schemes, as a full subcategory of the Zariski sheaves, are closed under finite limits and small coproducts. (Some discussion of these points can be found at the nForum, particularly in comment #18.)

Example

The category of affine schemes (opposite to the category of commutative rings with identity) is lextensive, but (perhaps contrary to geometric intuition) not infinitary lextensive. Some details may be found here.

Example

The category Cat is infinitary lextensive.

Example

The category Vect is not even finitely extensive.

Properties

Proposition

(in extensive categories connected objects are primitive under coproduct)
An object $X$ in an extensive category $\mathcal{C}$ is a connected object (in that the hom-functor $\mathcal{C}(X,-) \,\colon\, \mathcal{C} \to Set$ preserves coproducts) if and only if in any coproduct decomposition $X \simeq U + V$, exactly one of $U$, $V$ is not the initial object.

Proof

In one direction, assume that $X$ is connected and consider an isomorphism $X \xrightarrow{\sim} X_1 \sqcup X_2$ to a coproduct. By assumption of connectedness, this morphism factors through one of the summands, say through $X_1$, as shown in the bottom row of the following diagram:

Consider then the pullback of the total bottom morphism along the inclusion $i_2$ of the other summand. Since pullbacks of isomorphisms are isomorphisms, the resulting top left object must be (isomorphic to) $X_2$, as shown. On the other hand, by the pasting law this pullback factors into two pullback squares, as shown above. But the pullback on the right gives the initial object, since coproducts are disjoint in an extensive category (see above). This exhibits a morphisms $X_2 \to \varnothing$. But since the initial objects in extensive categories are strict initial objects, this must be an isomorphism, $X_2 \simeq \varnothing$. By the same argument, $X_1$ cannot be an initial object, since otherwise $X$ would be too, which it is not by assumption of connectedness (Rem. ). Hence we have shown that exactly one of the two summands in $X \simeq X_1 \sqcup X_2$ is initial.

In the other direction, assume that $X$ has non non-trivial coproduct decomposition and consider any morphism $f \colon X \to Y + Z$ into a coproduct. By extensivity, this implies (2) a coproduct decomposition $X = U + V$ with $U \coloneqq f^\ast(i_Y)$ and $V \coloneqq f^\ast(i_Z)$. But, by assumption, either $U$ or $V$ is initial, meaning that $X$ is isomorphic to either $V$ or $U$, respectively, so that $f$ factors through either $Y$ or $X$, respectively. In other words, $f$ belongs to exactly one of the two subsets $\hom(X, Y) \hookrightarrow \hom(X, Y + Z)$ or $\hom(X, Z) \hookrightarrow \hom(X, Y + Z)$.

Another useful fact is that any extensive category with finite products is distributive.

Related concepts

• extensive 2-category

• disjunctive (∞,1)-category

• extensive quantity

• objective number theory

• disjunctive logic

• Gaeta topos

References

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

• Aurelio Carboni, George Janelidze, Decidable (= separable) objects and morphisms in lextensive categories , JPAA 110 (1996) pp.210-240.

• Stephen Lack, Enrico M. Vitale, When do completion processes give rise to extensive categories? , JPAA 159 (2001) pp. 203-230 (doi:10.1016/S0022-4049(00)00060-8)

See also:

• Freyd, Lack, Lawvere et al: CATLIST discussion on extensive categories 1996