# nLab connected object

Contents

### Context

#### Compact objects

objects $d \in C$ such that $C(d,-)$ commutes with certain colimits

# Contents

## Idea

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

## Definitions

Let $\mathcal{C}$ be an extensive category.

###### Definition

An object $X$ of a category $\mathcal{C}$ is called connected if the hom-functor out of $X$

$\mathcal{C}(X, -) \,\colon\, \mathcal{C} \to Set \,.$
###### Remark

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

###### Remark

(initial object is not connected)
By Def. , the initial object $\varnothing$ of a category $\mathcal{C}$ is not connected:

Because, $\mathcal{C}(\varnothing, -) \,\colon\, \mathcal{C} \to Set$ is, by definition of initial objects, the constant functor with value the singleton $\ast \,\in\, Set$, so that for $X , Y \,\in\, \mathcal{C}$ we have

$\mathcal{C} \big( \varnothing ,\, X \sqcup Y \big) \;\simeq\; \ast \;\; \neq \;\; \ast \sqcup \ast \;\simeq\; \mathcal{C} (\varnothing,\, X) \sqcup \mathcal{C} (\varnothing,\, Y) \;\;\; \in \; Set \,.$

One may understand this as part of the pattern by which degenerate objects may be “too simple to be simple”. For example, also the empty space should not be considered connected, see this remark at connected space .

###### Remark

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

## Properties

### Characterization in terms of coproducts

Let $C$ be an infinitary extensive category, then

###### Proposition

An object $X$ of $C$ is connected, def. , if and only if the hom-functor $\hom(X, -) \colon C \to Set$ 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 above proof of Prop. 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.

###### Proposition

(in extensive categories connected objects are primitive under coproduct)
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

In one direction, assume that $X$ is connected and consider an isomorphism $X \xrightarrow{\sim} X_1 \sqcup X_2$ to a coproduct. By assumption of connectedness, this morphism factors through one of the summands, say through $X_1$, as shown in the bottom row of the following diagram:

Consider then the pullback of the total bottom morphism along the inclusion $i_2$ of the other summand. Since pullbacks of isomorphisms are isomorphisms, the resulting top left object must be (isomorphic to) $X_2$, as shown. On the other hand, by the pasting law this pullback factors into two pullback squares, as shown above. But the pullback on the right gives the initial object, since coproducts are disjoint in an extensive category (see here). This exhibits a morphisms $X_2 \to \varnothing$. But since the initial objects in extensive categories are strict initial objects, this must be an isomorphism, $X_2 \simeq \varnothing$. By the same argument, $X_1$ cannot be an initial object, since otherwise $X$ would be too, which it is not by assumption of connectedness (Rem. ). Hence we have shown that exactly one of the two summands in $X \simeq X_1 \sqcup X_2$ is initial.

In the other direction, assume that $X$ has non non-trivial coproduct decomposition and consider any morphism $f \colon X \to Y + Z$ into a coproduct. By extensivity, this implies (see here) a coproduct decomposition $X = U + V$ with $U \coloneqq f^\ast(i_Y)$ and $V \coloneqq f^\ast(i_Z)$. But, by assumption, either $U$ or $V$ is initial, meaning that $X$ is isomorphic to either $V$ or $U$, respectively, so that $f$ factors through either $Y$ or $X$, respectively. In other words, $f$ belongs to exactly one of the two subsets $\hom(X, Y) \hookrightarrow \hom(X, Y + Z)$ or $\hom(X, Z) \hookrightarrow \hom(X, Y + Z)$.

### 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 prop. 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

Last revised on November 7, 2021 at 23:55:58. See the history of this page for a list of all contributions to it.