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 below.

Properties

Characterization in terms of coproducts

Let $C$ be an infinitary extensive category, then

Theorem

An object $X$ of $C$ is connected, def. , 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. , 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 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 $i_U: U \hookrightarrow U + V$ is epic, in fact an epic equalizer of two maps $U + i_1, U + i_2: U + V \rightrightarrows U + V + V$. This means $i_U$ is an isomorphism; by disjointness of coproducts, this forces $V$ to be initial. Of course $U$ is not initial; otherwise $Y$ 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 $Top$.

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.

• connected graph

Related concepts

• connected topos, locally connected topos

• disjoint coproduct

• coproduct-preserving representable

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