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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.

\begin{remark} \label{InitialObjectIsNotConnected} (initial object is not connected) \linebreak By Def. 1, 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

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 .

\end{remark}

Properties

Characterization in terms of coproducts

Let $C$ be an infinitary extensive category, then

\begin{prop} \label{RespectForBinaryCoproductsIsSufficient} An object $X$ of $C$ is connected, def. 1, if and only if the hom-functor $\hom(X, -) \colon C \to Set$ preserves binary coproducts. \end{prop}

\begin{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

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

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

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 $\alpha \neq \beta$. \end{proof}

\begin{remark} This above proof of Prop. \ref{RespectForBinaryCoproductsIsSufficient} 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 $A$.) \end{remark}

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

\begin{proposition} \label{InExtensiveCategoryCOnnectedObjectsArePrimitiveUnderCoproduct} (in extensive categories connected objects are primitive under coproduct) \linebreak 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. \end{proposition}

\begin{proof} \label{ProofThatInExtensiveCategoryCOnnectedObjectsArePrimitiveUnderCoproduct} 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:

\begin{tikzcd} X_2 \ar[r] \ar[d] \ar[dr, phantom, \mbox{\tiny (pb)}] & \varnothing \ar[r] \ar[d] \ar[dr, phantom, \mbox{\tiny (pb)}] & X_2 \ar[d, i_2{right}] \ X \ar[r] \ar[rr, rounded corners, to path={ (yshift=-7pt\tikztostart.south) nodebelow{\scalebox{}} (yshift=-6pt\tikztotarget.south) (yshift=-0pt\tikztotarget.south) }] & X_1 \ar[r, i_1{below}] & X_1 \sqcup X_2 \mathrlap{\,.} \end{tikzcd} 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. \ref{InitialObjectIsNotConnected}). 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 $\hom(X, Z) \hookrightarrow \hom(X, Y + Z)$. \end{proof}

General properties

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

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

References

Called Connected objects were calledabstractly exclusively unaryabstractly exclusively unary objects in:

• Marta Bunge

  Marta Bunge, Categories of set valued functors, PhD thesis, University of Pennsylvania (1966) [pdf]

  , p. 32 of: Categories of set valued functors, PhD thesis, University of Pennsylvania (1966) [pdf]

• R. E. Hoffmann. A categorical concept of connectedness. Résumés Coll. Amiens. Cahiers. XIV-2 (1973). (pdf page 34)

• Reinhard Börger, Multicoreflective subcategories and coprime objects, Topology and its Applications 33.2 (1989): 127-142.