nLab
connected object

Contents

Idea

A connected object is a generalisation of the concept of connected space from Top to an arbitrary extensive category.

Definitions

Let CC be an extensive category, then:

Definition

An object XX of CC is connected if the representable functor

hom(X,):CSet hom(X, -) \colon C \to Set

preserves all coproducts.

Remark

By definition, hom(X,)hom(X,-) preserves binary coproducts if the canonically defined morphism hom(X,Y)+hom(X,Z)hom(X,Y+Z)hom(X,Y) + hom(X,Z) \to hom(X,Y + Z) is always a bijection.

Remark

By this definition, the initial object of CC is not in general connected (except for degenerate cases of CC); it is too simple to be simple. This matches the notion that the empty space should not be considered connected, discussed at connected space.

Remark

If CC is a infinitary extensive category then for XOb(C)X \in Ob(C) to be connected it is enough to require that hom(X,)hom(X,-) preserves binary coproducts. This is theorem below.

Properties

Characterization in terms of coproducts

Let CC be an infinitary extensive category, then

Theorem

An object XX of CC is connected, def. , if and only if hom(X,):CSet\hom(X, -): C \to Set preserves binary coproducts.

Proof

The “only if” is clear, so we just prove the “if”.

We first show that hom(X,)\hom(X, -) preserves the initial object 00. Indeed, if hom(X,)\hom(X, -) preserves the binary product X+0=XX + 0 = X, then the canonical map

hom(X,X)+hom(X,0)hom(X,X)\hom(X, X) + \hom(X, 0) \to \hom(X, X)

is a bijection of sets, where the restriction to hom(X,X)\hom(X, X) is also a bijection of sets id:hom(X,X)hom(X,X)id: \hom(X, X) \to \hom(X, X). This forces the set hom(X,0)\hom(X, 0) to be empty.

Now let {Y α:αA}\{Y_\alpha: \alpha \in A\} be a set of objects of CC. We are required to show that each map

f:X αY α f\colon X \to \sum_\alpha Y_\alpha

factors through a unique inclusion i α:Y α αY αi_\alpha: Y_\alpha \to \sum_\alpha Y_\alpha. By infinite extensivity, each pullback U αf *(Y α)U_\alpha \coloneqq f^\ast (Y_\alpha) exists and the canonical map αU αX\sum_\alpha U_\alpha \to X is an isomorphism. We will treat it as the identity.

Now all we need is to prove the following.

  • Claim: X=U αX = U_\alpha for exactly one α\alpha. For the others, U β=0U_\beta = 0.

Indeed, for each α\alpha, the identity map factors through one of the two summands in

id:XU α+ βαU β id\colon X \to U_\alpha + \sum_{\beta \neq \alpha} U_\beta

because hom(X,)\hom(X, -) preserves binary coproducts. In others words, either X=U αX = U_\alpha or X= βαU βX = \sum_{\beta \neq \alpha} U_\beta (and the other is 00). We cannot have U α=0U_\alpha = 0 for every α\alpha, for then X= αU αX = \sum_\alpha U_\alpha would be 00, contradicting the fact that hom(X,0)=0\hom(X, 0) = 0. So X=U αX = U_\alpha for at least one α\alpha. And no more than one α\alpha, since we have U αU β=0U_\alpha \cap U_\beta = 0 whenever αβ\alpha \neq \beta.

Remark

This proof is not constructive, as we have no way to construct a particular α\alpha such that X=U αX = U_\alpha. (It is constructive if Markov's principle applies to AA.)

From the proof above, we extract a result useful in its own right, giving an alternative definition of connected object.

Theorem

An object XX in an extensive category is connected, def. , if and only if in any coproduct decomposition XU+VX \simeq U + V, exactly one of UU, VV is not the initial object.

Proof

If XX is connected and X=U+VX = U + V is a coproduct decomposition, then the arrow id:XU+Vid \colon X \to U + V, factors through one of the coproduct inclusions of i U,i V:U,VU+Vi_U, i_V \colon U, V \hookrightarrow U + V. If it factors through say UU, then the subobject i Ui_U is all of XX, and VV is forced to be initial by disjointness of coproducts.

Turning now to the if direction, suppose f:XY+Zf \colon X \to Y + Z is a map, and put U=f *(i Y)U = f^\ast(i_Y), V=f *(i Z)V = f^\ast(i_Z). By extensivity, we have a coproduct decomposition X=U+VX = U + V. One of UU, VV is initial, say VV, and then we have X=UX = U, meaning that ff factors through YY, and uniquely so since i Yi_Y is monic. Hence ff belongs to (exactly) one of the two subsets hom(X,Y)hom(X,Y+Z)\hom(X, Y) \hookrightarrow \hom(X, Y + Z), hom(X,Z)hom(X,Y+Z)\hom(X, Z) \hookrightarrow \hom(X, Y + Z).

General properties

Proposition

A colimit of connected objects over a connected diagram is itself a connected object.

Proof

Because coproducts in SetSet commute with limits of connected diagrams.

Proposition

If XOb(C)X \in Ob(C) is connected and XYX \to Y is an epimorphism, then YY is connected.

Proof

Certainly YY is not initial, because initial objects in extensive categories are strict. Suppose Y=U+VY = U + V (see theorem above), so that we have an epimorphism XU+VX \to U + V. By connectedness of XX, this epi factors through one of the summands, say UU. But then the inclusion i U:UU+Vi_U: U \hookrightarrow U + V is epic, in fact an epic equalizer of two maps U+i 1,U+i 2:U+VU+V+VU + i_1, U + i_2: U + V \rightrightarrows U + V + V. This means i Ui_U is an isomorphism; by disjointness of coproducts, this forces VV to be initial. Of course UU is not initial; otherwise YY would be initial.

Remark

It need not be the case that products of connected objects are connected. For example, in the topos \mathbb{Z}-Set, the product ×\mathbb{Z} \times \mathbb{Z} decomposes as a countable coproduct of copies of \mathbb{Z}. (For more on this topic, see also cohesive topos.)

We do have the following partial result, generalizing the case of TopTop.

Theorem

Suppose CC is a cocomplete \infty-extensive category with finite products. Assume that the terminal object is a separator, and that Cartesian product functors X×():CCX \times (-) : C \to C preserve epimorphisms. Then a product of finitely many connected objects is itself connected.

Proof

In the first place, 11 is connected. For suppose 1=U+V1 = U + V where UU is not initial. The two coproduct inclusions UU+UU \to U + U are distinct by disjointness of sums. Since 11 is a separator, there must be a map 1U1 \to U separating these inclusions. We then conclude U1U \cong 1, and then V0V \cong 0 by disjointness of sums.

Now let XX and YY be connected. The two inclusions i 1,i 2:XX+Xi_1, i_2: X \to X + X are distinct, so there exists a point a:1Xa \colon 1 \to X separating them. For each y:1Yy \colon 1 \to Y, the +-shaped object T y=(X×y)(a×Y)T_y = (X \times y) \cup (a \times Y) is connected (we define T yT_y to be a sum of connected objects X×yX \times y, a×Ya \times Y amalgamated over the connected object a×y=1×1=1a \times y = 1 \times 1 = 1, i.e., to be a connected colimit of connected objects). We have a map

ϕ: y:1YT yX×Y\phi: \bigcup_{y \colon 1 \to Y} T_y \to X \times Y

where the union is again a sum of connected objects T yT_y amalgamated over a×Ya \times Y, so this union is connected. The map ϕ\phi is epic, because the evident map

y:1YX×yX× y:1Y1X×Y\sum_{y \colon 1 \to Y} X \times y \cong X \times \sum_{y \colon 1 \to Y} 1 \to X \times Y

is epic (by the assumptions that 11 is a separator and X×X \times - preserves epis), and this map factors through ϕ\phi. It follows that the codomain X×YX \times Y of ϕ\phi is also connected.

Remark

The same method of proof shows that for an arbitrary family of connected spaces {X α} αA\{X_\alpha\}_{\alpha \in A}, the connected component of a point x=(x α)x = (x_\alpha) in the product space X= αAX αX = \prod_{\alpha \in A} X_\alpha contains at least all those points which differ from xx in at most finitely many coordinates. However, the set of such points is dense in αAX α\prod_{\alpha \in A} X_\alpha, so αAX α\prod_{\alpha \in A} X_\alpha must also be connected.

Examples

Last revised on January 17, 2018 at 06:33:40. See the history of this page for a list of all contributions to it.