nLab
connected object

Contents

Contents

Idea

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

Definitions

Let π’ž\mathcal{C} be an extensive category.

Definition

An object XX of a category π’ž\mathcal{C} is called connected if the hom-functor out of XX

π’ž(X,βˆ’):π’žβ†’Set. \mathcal{C}(X, -) \,\colon\, \mathcal{C} \to Set \,.

preserves all coproducts.

Remark

By definition, π’ž(X,βˆ’)\mathcal{C}(X,-) preserves binary coproducts if the canonically defined morphism π’ž(X,Y)+π’ž(X,Z)β†’π’ž(X,Y+Z)\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 βˆ…\varnothing of a category π’ž\mathcal{C} is not connected:

Because, π’ž(βˆ…,βˆ’):π’žβ†’Set\mathcal{C}(\varnothing, -) \,\colon\, \mathcal{C} \to Set is, by definition of initial objects, the constant functor with value the singleton *∈Set\ast \,\in\, Set, so that for X,Yβˆˆπ’žX , Y \,\in\, \mathcal{C} we have

π’ž(βˆ…,XβŠ”Y)≃*β‰ *βŠ”*β‰ƒπ’ž(βˆ…,X)βŠ”π’ž(βˆ…,Y)∈Set. \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 β€œtoo simple to be simple”. For example, also the empty space should not be considered connected, see this remark at connected space .

Remark

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

Properties

Characterization in terms of coproducts

Let CC be an infinitary extensive category, then

Proposition

An object XX of CC is connected, def. , if and only if the hom-functor hom(X,βˆ’):Cβ†’Set\hom(X, -) \colon C \to Set preserves binary coproducts.

Proof

The β€œonly if” is clear, so we just prove the β€œif”.

We first show that hom(X,βˆ’)\hom(X, -) preserves the initial object 00. Indeed, if hom(X,βˆ’)\hom(X, -) preserves the binary product X+0=XX + 0 = X, then the canonical map

hom(X,X)+hom(X,0)β†’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)β†’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:α∈A}\{Y_\alpha: \alpha \in A\} be a set of objects of CC. We are required to show that each map

f:Xβ†’βˆ‘ Ξ±Y Ξ± f\colon X \to \sum_\alpha Y_\alpha

factors through a unique inclusion i Ξ±:Y Ξ±β†’βˆ‘ Ξ±Y Ξ±i_\alpha: Y_\alpha \to \sum_\alpha Y_\alpha. By infinite extensivity, each pullback U α≔f *(Y Ξ±)U_\alpha \coloneqq f^\ast (Y_\alpha) exists and the canonical map βˆ‘ Ξ±U Ξ±β†’X\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 Ξ±X = U_\alpha for exactly one Ξ±\alpha. For the others, U Ξ²=0U_\beta = 0.

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

id:Xβ†’U Ξ±+βˆ‘ Ξ²β‰ Ξ±U Ξ² id\colon X \to U_\alpha + \sum_{\beta \neq \alpha} U_\beta

because hom(X,βˆ’)\hom(X, -) preserves binary coproducts. In others words, either X=U Ξ±X = U_\alpha or X=βˆ‘ Ξ²β‰ Ξ±U Ξ²X = \sum_{\beta \neq \alpha} U_\beta (and the other is 00). We cannot have U Ξ±=0U_\alpha = 0 for every Ξ±\alpha, for then X=βˆ‘ Ξ±U Ξ±X = \sum_\alpha U_\alpha would be 00, contradicting the fact that hom(X,0)=0\hom(X, 0) = 0. So X=U Ξ±X = U_\alpha for at least one Ξ±\alpha. And no more than one Ξ±\alpha, since we have U α∩U Ξ²=0U_\alpha \cap U_\beta = 0 whenever Ξ±β‰ Ξ²\alpha \neq \beta.

Remark

This above proof of Prop. is not constructive, as we have no way to construct a particular Ξ±\alpha such that X=U Ξ±X = 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.

Proposition

(in extensive categories connected objects are primitive under coproduct)
An object XX in an extensive category is connected, def. , if and only if in any coproduct decomposition X≃U+VX \simeq U + V, exactly one of UU, VV is not the initial object.

Proof

In one direction, assume that XX is connected and consider an isomorphism Xβ†’βˆΌX 1βŠ”X 2X \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 1X_1, as shown in the bottom row of the following diagram:

Consider then the pullback of the total bottom morphism along the inclusion i 2i_2 of the other summand. Since pullbacks of isomorphisms are isomorphisms, the resulting top left object must be (isomorphic to) X 2X_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 2β†’βˆ…X_2 \to \varnothing. But since the initial objects in extensive categories are strict initial objects, this must be an isomorphism, X 2β‰ƒβˆ…X_2 \simeq \varnothing. By the same argument, X 1X_1 cannot be an initial object, since otherwise XX 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≃X 1βŠ”X 2X \simeq X_1 \sqcup X_2 is initial.

In the other direction, assume that XX has non non-trivial coproduct decomposition and consider any morphism f:Xβ†’Y+Zf \colon X \to Y + Z into a coproduct. By extensivity, this implies (see here) a coproduct decomposition X=U+VX = U + V with U≔f *(i Y)U \coloneqq f^\ast(i_Y) and V≔f *(i Z)V \coloneqq f^\ast(i_Z). But, by assumption, either UU or VV is initial, meaning that XX is isomorphic to either VV or UU, respectively, so that ff factors through either YY or XX, respectively. In other words, ff belongs to exactly one of the two subsets hom(X,Y)β†ͺhom(X,Y+Z)\hom(X, Y) \hookrightarrow \hom(X, Y + Z) or hom(X,Z)β†ͺhom(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 SetSet commute with limits of connected diagrams.

Proposition

If X∈Ob(C)X \in Ob(C) is connected and Xβ†’YX \to Y is an epimorphism, then YY is connected.

Proof

Certainly YY is not initial, because initial objects in extensive categories are strict. Suppose Y=U+VY = U + V (see prop. above), so that we have an epimorphism Xβ†’U+VX \to U + V. By connectedness of XX, this epi factors through one of the summands, say UU. But then the inclusion i U:Uβ†ͺU+Vi_U: U \hookrightarrow U + V is epic, in fact an epic equalizer of two maps U+i 1,U+i 2:U+V⇉U+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.

Remark

It need not be the case that products of connected objects are connected. For example, in the topos β„€\mathbb{Z}-Set, the product β„€Γ—β„€\mathbb{Z} \times \mathbb{Z} decomposes as a countable coproduct of copies of β„€\mathbb{Z}. (For more on this topic, see also cohesive topos.)

We do have the following partial result, generalizing the case of TopTop.

Theorem

Suppose CC is a cocomplete ∞\infty-extensive category with finite products. Assume that the terminal object is a separator, and that Cartesian product functors XΓ—(βˆ’):Cβ†’CX \times (-) : C \to C preserve epimorphisms. Then a product of finitely many connected objects is itself connected.

Proof

In the first place, 11 is connected. For suppose 1=U+V1 = U + V where UU is not initial. The two coproduct inclusions U→U+UU \to U + U are distinct by disjointness of sums. Since 11 is a separator, there must be a map 1→U1 \to U separating these inclusions. We then conclude U≅1U \cong 1, and then V≅0V \cong 0 by disjointness of sums.

Now let XX and YY be connected. The two inclusions i 1,i 2:Xβ†’X+Xi_1, i_2: X \to X + X are distinct, so there exists a point a:1β†’Xa \colon 1 \to X separating them. For each y:1β†’Yy \colon 1 \to Y, the +-shaped object T y=(XΓ—y)βˆͺ(aΓ—Y)T_y = (X \times y) \cup (a \times Y) is connected (we define T yT_y to be a sum of connected objects XΓ—yX \times y, aΓ—Ya \times Y amalgamated over the connected object aΓ—y=1Γ—1=1a \times y = 1 \times 1 = 1, i.e., to be a connected colimit of connected objects). We have a map

Ο•:⋃ y:1β†’YT yβ†’XΓ—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 aΓ—Ya \times Y, so this union is connected. The map Ο•\phi is epic, because the evident map

βˆ‘ y:1β†’YXΓ—yβ‰…XΓ—βˆ‘ y:1β†’Y1β†’XΓ—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 XΓ—βˆ’X \times - preserves epis), and this map factors through Ο•\phi. It follows that the codomain XΓ—YX \times Y of Ο•\phi is also connected.

Remark

The same method of proof shows that for an arbitrary family of connected spaces {X α} α∈A\{X_\alpha\}_{\alpha \in A}, the connected component of a point x=(x α)x = (x_\alpha) in the product space X=∏ α∈AX 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 ∏ α∈AX α\prod_{\alpha \in A} X_\alpha, so ∏ α∈AX α\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.