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, then:

Definition

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

$$hom(X, -) \colon C \to Set$$

preserves all coproducts.

Remark

By definition, $hom(X,-)$ preserves binary coproducts if the canonically defined morphism $hom(X,Y) + hom(X,Z) \to hom(X,Y + Z)$ 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 $X \in Ob(C)$ to be connected it is enough to require that $hom(X,-)$ preserves binary coproducts. This is theorem 1 below.

Properties

Characterization in terms of coproducts

Let $C$ be an infinitary extensive category, then

Theorem

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

Proof

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

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

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

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

$$f\colon X \to \sum_\alpha Y_\alpha$$

factors through a unique inclusion $i_\alpha: Y_\alpha \to \sum_\alpha Y_\alpha$. By infinite extensivity, each pullback $U_\alpha \coloneqq f^\ast (Y_\alpha)$ 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

$$id\colon X \to U_\alpha + \sum_{\beta \neq \alpha} U_\beta$$

because $\hom(X, -)$ preserves binary coproducts. In others words, either $X = U_\alpha$ or $X = \sum_{\beta \neq \alpha} U_\beta$ (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 $\hom(X, 0) = 0$. 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 $\alpha \neq \beta$.

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 $id \colon X \to U + V$, factors through one of the coproduct inclusions of $i_U, i_V \colon U, V \hookrightarrow U + V$. 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 $f \colon X \to Y + Z$ is a map, and put $U = f^\ast(i_Y)$, $V = f^\ast(i_Z)$. 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 $\hom(X, Y) \hookrightarrow \hom(X, Y + Z)$, $\hom(X, Z) \hookrightarrow \hom(X, Y + Z)$.

General properties

Proposition

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

Proof

Because coproducts in $Set$ commute with limits of connected diagrams.

Proposition

If $X \in Ob(C)$ 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 $Top$.

Theorem

Suppose $C$ is a cocomplete $\infty$-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 $a \colon 1 \to X$ separating them. For each $y \colon 1 \to Y$, 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

$$\phi: \bigcup_{y \colon 1 \to Y} T_y \to X \times Y$$

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

$$\sum_{y \colon 1 \to Y} X \times y \cong X \times \sum_{y \colon 1 \to Y} 1 \to X \times Y$$

is epic (by the assumptions that $1$ is a separator and $X \times -$ preserves epis), and this map factors through $\phi$. It follows that the codomain $X \times Y$ of $\phi$ is also connected.

Remark

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

Examples

• Connected objects in Top are precisely connected topological spaces.

• For a group $G$, connected objects in the category $G Set$ 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 $G Set$, permutation representations of a group $G$.)

• A scheme is connected as a scheme (which by definition means that its underlying topological space is connected) if and only if it is connected as an object of the category of schemes. See here.

Related concepts

• connected topos, locally connected topos

• disjoint coproduct

• coproduct-preserving representable

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