connected object



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


Let CC be an extensive category, then:


An object XX of CC is connected if the representable functor

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

preserves all coproducts.


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.


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.


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 1 below.


Characterization in terms of coproducts

Let CC be an infinitary extensive category, then


An object XX of CC is connected, def. 1, if and only if hom(X,):CSet\hom(X, -): C \to Set preserves binary coproducts.


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.


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.


An object XX in an extensive category is connected, def. 1, if and only if in any coproduct decomposition XU+VX \simeq U + V, exactly one of UU, VV is not the initial object.


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


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


Because coproducts in SetSet commute with limits of connected diagrams.


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


Certainly YY is not initial, because initial objects in extensive categories are strict. Suppose Y=U+VY = U + V (see theorem 2 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.


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.


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.


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.


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.


Revised on January 17, 2018 06:33:40 by Urs Schreiber (