nLab
connected object

Idea

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

Definitions

Let $C$ be an extensive category.

Definition

An object $X$ of $C$ is connected if the representable functor

$$\mathrm{hom}(X,-):C\to \mathrm{Set}$$hom(X, -) \colon C \to Set

preserves all coproducts.

Remark

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

Remark

By this definition, the initial object of $C$ is not in general connected (except for degenerate cases of $C$); 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 $C$ is a infinitary extensive category then for $X\in \mathrm{Ob}(C)$ to be connected it is enough to require that $\mathrm{hom}(X,-)$ it preserves binary coproducts. This is theorem 1 below.

Properties

Characterization in terms of coproducts

Let $C$ be an infinitary extensive category.

Theorem

Then an object $X$ of $C$ is connected, def. 1, if and only if $\mathrm{hom}(X,-):C\to \mathrm{Set}$ preserves binary coproducts.

Proof

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

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

$$\mathrm{hom}(X,X)+\mathrm{hom}(X,0)\to \mathrm{hom}(X,X)$$\hom(X, X) + \hom(X, 0) \to \hom(X, X)

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

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

$$f:X\to \sum _{\alpha }{Y}_{\alpha }$$f\colon X \to \sum_\alpha Y_\alpha

factors through a unique inclusion $i_{\alpha }:Y_{\alpha }\to {\sum }_{\alpha }{Y}_{\alpha }$. By infinite extensivity, each pullback $U_{\alpha }≔f^{*}(Y_{\alpha })$ exists and the canonical map ${\sum }_{\alpha }{U}_{\alpha }\to X$ is an isomorphism. We will treat it as the identity.

Now all we need is to prove the following.

• Claim: $X=U_{\alpha }$ for exactly one $\alpha$. For the others, $U_{\beta }=0$.

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

$$\mathrm{id}:X\to U_{\alpha }+\sum _{\beta \ne \alpha }{U}_{\beta }$$id\colon X \to U_\alpha + \sum_{\beta \neq \alpha} U_\beta

because $\mathrm{hom}(X,-)$ preserves binary coproducts. In others words, either $X=U_{\alpha }$ or $X={\sum }_{\beta \ne \alpha }{U}_{\beta }$ (and the other is $0$). We cannot have $U_{\alpha }=0$ for every $\alpha$, for then $X={\sum }_{\alpha }{U}_{\alpha }$ would be $0$, contradicting the fact that $\mathrm{hom}(X,0)=0$. So $X=U_{\alpha }$ for at least one $\alpha$. And no more than one $\alpha$, since we have $U_{\alpha }\cap U_{\beta }=0$ whenever $\alpha \ne \beta$.

Remark

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

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

Theorem

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

Proof

If $X$ is connected and $X=U+V$ is a coproduct decomposition, then the arrow $\mathrm{id}:X\to U+V$, factors through one of the coproduct inclusions of $i_U,i_V:U,V\hookrightarrow U+V$. If it factors through say $U$, then the subobject $i_U$ is all of $X$, and $V$ is forced to be initial by disjointness of coproducts.

Turning now to the if direction, suppose $f:X\to Y+Z$ is a map, and put $U=f^{*}(i_Y)$, $V=f^{*}(i_Z)$. By extensivity, we have a coproduct decomposition $X=U+V$. One of $U$, $V$ is initial, say $V$, and then we have $X=U$, meaning that $f$ factors through $Y$, and uniquely so since $i_Y$ is monic. Hence $f$ belongs to (exactly) one of the two subsets $\mathrm{hom}(X,Y)\hookrightarrow \mathrm{hom}(X,Y+Z)$, $\mathrm{hom}(X,Z)\hookrightarrow \mathrm{hom}(X,Y+Z)$.

General properties

Proposition

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

Proof

Because coproducts in $\mathrm{Set}$ commute with limits of connected diagrams.

Proposition

If $X\in \mathrm{Ob}(C)$ is connected and $X\to Y$ is an epimorphism, then $Y$ is connected.

Proof

Certainly $Y$ is not initial, because initial objects in extensive categories are strict. Suppose $Y=U+V$ (see theorem 2 above), so that we have an epimorphism $X\to U+V$. By connectedness of $X$, this epi factors through one of the summands, say $U$. But then the inclusion $U\hookrightarrow U+V$ is epi. This forces $V$ to 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 $\mathrm{Top}$.

Theorem

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

Proof

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

Now let $X$ and $Y$ be connected. The two inclusions $i_1,i_2:X\to X+X$ are distinct, so there exists a point $a:1\to X$ separating them. For each $y:1\to Y$, the +-shaped object $T_y=(X\times y)\cup (a\times Y)$ is connected (we define $T_y$ to be a sum of connected objects $X\times y$, $a\times Y$ amalgamated over the connected object $a\times y=1\times 1=1$, i.e., to be a connected colimit of connected objects). We have a map

$$\varphi :\bigcup _{y:1\to Y}{T}_y\to X\times Y$$\phi: \bigcup_{y \colon 1 \to Y} T_y \to X \times Y

where the union is again a sum of connected objects $T_y$ amalgamated over $a\times Y$, so this union is connected. The map $\varphi$ is epic, because the evident map

$$\sum _{y:1\to Y}X\times y\cong X\times \sum _{y:1\to Y}1\to X\times Y$$\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 $1$ is a separator and $X\times -$ preserves epis), and this map factors through $\varphi$. It follows that the codomain $X\times Y$ of $\varphi$ is also connected.

Examples

• Connected objects in Top are precisely connected topological spaces.

• For a group $G$, connected objects in the category $G\mathrm{Set}$ of permutation representations are precisely the (inhabited) transitive $G$-sets.

• Objects in a locally connected topos are coproducts of connected objects.

  (This includes categories such as $G\mathrm{Set}$, permutation representations of a group $G$.)

Related concepts

• connected topos, locally connected topos

• disjoint coproduct

• coproduct-preserving representable

• n-connected object in an (∞,1)-topos