Connected Objects in Extensive Categories

Let $\mathcal{C}$$\mathcal{C}$ be a locally small extensive category that possesses a terminal object $1$$1$.

For an object $X\in \mathcal{C}$$X \in \mathcal{C}$, the following three conditions are equivalent:

1. The hom-functor ${\mathrm{Hom}}_{\mathcal{C}}(X,-):\mathcal{C}\to \mathsf{Set}$$\operatorname{Hom}_{\mathcal{C}}(X, -) : \mathcal{C} \to \mathsf{Set}$ preserves finite coproducts.

2. The canonical map

   $${\mathrm{Hom}}_{\mathcal{C}}(X,1)+{\mathrm{Hom}}_{\mathcal{C}}(X,1)\stackrel{\cong }{\to }{\mathrm{Hom}}_{\mathcal{C}}(X,1+1)$$

   is a bijection (which forces $X$$X$ to be non-initial).

3. $X$$X$ is not initial and cannot be written as a coproduct $X\cong X_1+X_2$$X \cong X_1 + X_2$ with both $X_1,X_2$$X_1, X_2$ non-initial.

An object satisfying $(1)\sim (3)$$(1)\sim(3)$ is called connected. Examples: $\mathsf{Top}$$\mathsf{Top}$, $\mathsf{Sch}$$\mathsf{Sch}$, any topos (e.g. $\mathsf{Quiver}$$\mathsf{Quiver}$, $G$$G$‑$\mathsf{Set}$$\mathsf{Set}$).

In $\mathsf{Top}$$\mathsf{Top}$, $\mathsf{Sch}$$\mathsf{Sch}$, and $\mathsf{Quiver}$$\mathsf{Quiver}$, it is just connected; in $G\text{-}\mathsf{Set}$$G\text{-}\mathsf{Set}$, it is a transitive $G$$G$-set.

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Proof of the equivalences

We assume throughout that $\mathcal{C}$$\mathcal{C}$ is extensive and has a terminal object $1$$1$. Coproduct injections are denoted $i_1:A\to A+B$$i_1 : A \to A+B$ and $i_2:B\to A+B$$i_2 : B \to A+B$.

$(1)\Rightarrow (3)$$(1) \Rightarrow (3)$

Suppose $\mathrm{Hom}(X,-)$$\operatorname{Hom}(X,-)$ preserves finite coproducts.
Then $\mathrm{Hom}(X,0)\cong \varnothing$$\operatorname{Hom}(X,0) \cong \varnothing$ (empty coproduct), so $X$$X$ is not initial.

Assume, for contradiction, that $X\cong X_1+X_2$$X \cong X_1 + X_2$ with $X_1,X_2\ncong 0$$X_1,X_2 \not\cong 0$.
The identity ${\mathrm{id}}_X$$\operatorname{id}_X$ corresponds under the isomorphism $X\cong X_1+X_2$$X \cong X_1+X_2$ to a morphism $f:X\to X_1+X_2$$f: X \to X_1+X_2$.
Since $\mathrm{Hom}(X,-)$$\operatorname{Hom}(X,-)$ preserves the coproduct,

$$\mathrm{Hom}(X,X_1+X_2)\cong \mathrm{Hom}(X,X_1)\bigsqcup \mathrm{Hom}(X,X_2),$$

the morphism $f$$f$ must belong to exactly one of the two summands.
Without loss of generality assume it belongs to the first; thus there exists $u:X\to X_1$$u: X \to X_1$ such that $f=i_1\circ u$$f = i_1 \circ u$.

Now pull back the coproduct inclusions along $f$$f$:

$$\begin{array}{ccc}{P}_1& \stackrel{}{\to }& X_1\\ \downarrow & & \downarrow i_1& \\ X& \stackrel{f}{\to }& X_1+X_2\\ \uparrow & & \uparrow i_2& \\ P_2& \stackrel{}{\to }& X_2\end{array}$$

Extensivity gives $X\cong P_1+P_2$$X \cong P_1 + P_2$.
Because $f$$f$ factors through $i_1$$i_1$, the top square is a pullback of an isomorphism along $i_1$$i_1$, hence $P_1\cong X$$P_1 \cong X$. By disjointness of the coproduct, the pullback of $i_1$$i_1$ and $i_2$$i_2$ is the initial object $0$$0$; consequently $P_2\cong 0$$P_2 \cong 0$. Thus $X\cong X+0$$X \cong X + 0$, forcing $X_2\cong 0$$X_2 \cong 0$ (by comparing the two decompositions), contradicting the assumption that $X_2$$X_2$ is non-initial.

Hence no non-trivial decomposition exists — i.e. condition $(3)$$(3)$ holds.

$(3)\Rightarrow (1)$$(3) \Rightarrow (1)$

Assume $X$$X$ is non-initial and cannot be split into two non-initial summands.
We show the natural map

$$\underset{j\in J}{⨆}\mathrm{Hom}(X,A_j)⟶\mathrm{Hom}(X,\coprod _{j\in J}{A}_j)$$

is a bijection for every finite family $(A_j)$$(A_j)$. By induction it suffices to treat binary coproducts $A+B$$A+B$.

Take an arbitrary $f:X\to A+B$$f: X \to A+B$.
Pull back the injections along $f$$f$:

$$\begin{array}{ccc}{X}_1& \stackrel{}{\to }& A\\ \downarrow & & \downarrow i_1& \\ X& \stackrel{f}{\to }& A+B\\ \uparrow & & \uparrow i_2& \\ X_2& \stackrel{}{\to }& B\end{array}$$

Extensivity yields $X\cong X_1+X_2$$X \cong X_1 + X_2$.

By condition $(3)$$(3)$, one of $X_1,X_2$$X_1, X_2$ must be initial. If $X_2\cong 0$$X_2 \cong 0$, then $f$$f$ factors uniquely through $i_1:A\to A+B$$i_1 : A \to A+B$ (the map from $0$$0$ is unique and the universal property of the coproduct collapses onto the remaining summand).
Hence every morphism $X\to A+B$$X \to A+B$ lands entirely in one summand, which is precisely the statement that

$$\mathrm{Hom}(X,A)\bigsqcup \mathrm{Hom}(X,B)\stackrel{\cong }{\to }\mathrm{Hom}(X,A+B)$$

is an isomorphism. Together with $\mathrm{Hom}(X,0)\cong \varnothing$$\operatorname{Hom}(X,0) \cong \varnothing$ (true for any non-initial $X$$X$ because $0$$0$ is strict in an extensive category), we obtain preservation of all finite coproducts, i.e. $(1)$$(1)$.

$(3)\iff (2)$$(3) \Leftrightarrow (2)$

Decompositions of $X$$X$ are in bijection with morphisms into $1+1$$1+1$:

• Decomposition $\mapsto$$\mapsto$ map:
  If $X\cong X_1+X_2$$X \cong X_1 + X_2$, send it to the copairing $X_1\to 1$$X_1 \to 1$, $X_2\to 1$$X_2 \to 1$, yielding $f:X\to 1+1$$f: X \to 1+1$.

• Map $\mapsto$$\mapsto$ decomposition:
  Given $f:X\to 1+1$$f: X \to 1+1$, pull back ${\mathrm{\top }}_1:1\to 1+1$$\top_1 : 1 \to 1+1$ and ${\mathrm{\top }}_2:1\to 1+1$$\top_2 : 1 \to 1+1$ to obtain $P_1,P_2$$P_1, P_2$ with $X\cong P_1+P_2$$X \cong P_1 + P_2$.

These constructions are mutually inverse, establishing

$$\{\text{decompositions }X\cong X_1+X_2\}⟷\mathrm{Hom}(X,1+1).$$

The two summands on the left of the canonical map

$$\mathrm{Hom}(X,1)+\mathrm{Hom}(X,1)⟶\mathrm{Hom}(X,1+1)$$

correspond exactly to the trivial decompositions $X\cong X+0$$X \cong X + 0$ and $X\cong 0+X$$X \cong 0 + X$.

• If $X$$X$ satisfies $(3)$$(3)$, every decomposition is trivial, so the map is bijective and $X$$X$ is non-initial, giving $(2)$$(2)$.

• If $(2)$$(2)$ holds, the bijection forces every map $X\to 1+1$$X \to 1+1$ to come from a trivial decomposition, hence $X$$X$ has no non-trivial decomposition. Moreover $X$$X$ cannot be initial (otherwise left side would be $1+1$$1+1$, right side a singleton). Thus $(3)$$(3)$ holds.

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Image of a connected object is connected

Assume now that $\mathcal{C}$$\mathcal{C}$ has an (epi, mono) factorization system.

Proposition. If $X$$X$ is connected and $f:X\to Y$$f: X \to Y$ is any morphism, then $\mathrm{Im}(f)$$\operatorname{Im}(f)$ is connected.

Proof. Factor $f$$f$ as $X\stackrel{e}{\to }\mathrm{Im}f\stackrel{m}{\to }Y$$X \xrightarrow{e} \operatorname{Im}f \xrightarrow{m} Y$ with $e$$e$ epic and $m$$m$ monic. Suppose $\mathrm{Im}f$$\operatorname{Im}f$ were not connected. Then there exist non-initial $A,B$$A,B$ and an isomorphism $h:\mathrm{Im}f\stackrel{\cong }{\to }A+B$$h: \operatorname{Im}f \xrightarrow{\cong} A+B$. Define $g=h\circ e:X\to A+B$$g = h \circ e : X \to A+B$ and pull back the coproduct injections along $g$$g$:

$$\begin{array}{ccc}{X}_A& \stackrel{}{\to }& A\\ \downarrow & & \downarrow i_1& \\ X& \stackrel{g}{\to }& A+B\\ \uparrow & & \uparrow i_2& \\ X_B& \stackrel{}{\to }& B\end{array}$$

Extensivity yields $X\cong X_A+X_B$$X \cong X_A + X_B$.

Claim: Both $X_A$$X_A$ and $X_B$$X_B$ are non-initial. If $X_A\cong 0$$X_A \cong 0$, then $g$$g$ factors through $i_2$$i_2$, implying $\mathrm{Im}(g)\subseteq B$$\operatorname{Im}(g) \subseteq B$. But $g=h\circ e$$g = h \circ e$ and $h$$h$ is an isomorphism, so $\mathrm{Im}(g)\cong \mathrm{Im}(e)=\mathrm{Im}f\cong A+B$$\operatorname{Im}(g) \cong \operatorname{Im}(e) = \operatorname{Im}f \cong A+B$. This forces $A\cong 0$$A \cong 0$ (since the coproduct summand $A$$A$ must be contained in $B$$B$, which is impossible in an extensive category unless $A$$A$ is initial), contradicting $A\ncong 0$$A \not\cong 0$. Similarly $X_B\cong 0$$X_B \cong 0$ would force $B\cong 0$$B \cong 0$, a contradiction.

Thus $X\cong X_A+X_B$$X \cong X_A + X_B$ is a non-trivial coproduct decomposition of $X$$X$, contradicting the assumption that $X$$X$ is connected. Hence $\mathrm{Im}f$$\operatorname{Im}f$ must be connected. $◻$$\square$