In an extensive category, the following conditions are equivalent for an object.


1. In a coproduct decomposition, exactly one of the summands is initial;
2. The covariant functor it represents preserves coproducts.


An object satisfying these equivalent conditions is called *connected*.


Now consider the functor $\mathsf C\overset{H}{\longleftarrow} \mathsf{Set}$ given by taking copowers of a fixed terminal object. Say an object $X$ of $\mathsf C$ *has connected components* if there exists a universal arrow $\eta\_X:X\to H(\Pi\_0 X)$ from $X$ to $H$.


The universal property of this universal arrow means $\Pi\_0(X)\cong \mathbf 1$ is equivalent to the covariant $\mathrm{Hom}(X,-)$ preserving coproducts of *terminal* objects. Thus it's clear connected objects have a single connected component.


Unfortunately, I am struggling with the converse. Although I don't expect $\Pi\_0$ to exist globally as a left adjoint to $H$ (e.g not all topological spaces are the coproducts of their connected components), it still seems reasonable to expect that $\Pi\_0(X)\cong \mathbf 1$ should *imply* $X$ is connected in any extensive category.


But that would mean that $\mathrm{Hom}(X,-)$ preserving coproducts of terminal objects is *equivalent* to $\mathrm{Hom}(X,-)$ preserving arbitrary coproducts. I think I have proved this given $\Pi\_0\dashv H$, but I am not as confident about the general case:


It seems that in an extensive category $\mathrm{Hom}(X,\bf 2)$ is in bijection with binary coproduct decompositions of $X$. If $\mathrm{Hom}(X,-)$ preserves coproducts of terminal objects then $\mathrm{Hom}(X,\mathbf 2)\cong \mathbf 2$ which must be the two trivial coproduct decompositions $\mathbf 0\amalg X\cong X\cong X\amalg \mathbf 0$. This seems to imply $X$ is connected.


Is the latter argument correct?



Connectedness in extensive categories in terms of set of connected components
I wouldn't say that there is a bijection between $\mathrm{Hom}(X,2)$ and binary coproduct decompositions of $X$ since the former is a set and the latter is a category. If we ignore this fact, then we can show that the natural function $\coprod\limits\_{A,B \in \mathrm{Ob(C)}} A \amalg B \simeq X \to \mathrm{Hom}(X,2)$ is surjective, so there is a bijection if we define the set of coproduct decompositions in a certain way, but I don't know how to use it to prove this fact. I can give another argument instead.


Let's say that we have a decomposition $X \simeq A \amalg B$. Let $f : X \to 2$ be the map $X \to A + B \to 2$. Since the obvious map $2 \to \mathrm{Hom}(X,2)$ is a bijection, $f$ factors through either $i\_0 : 1 \to 2$ or $i\_1 : 1 \to 2$. Let's say it factors through $i\_0$ (the other case is similar). It follows that the map $A + B \to 2$ also factors through $i\_0$. Then the map $B \to A + B \to 2$ equals to $B \to A + B \to 1 \overset{i\_0}\to 2$ and it also equals to $B \to 1 \overset{i\_1}\to 2$ by definition. Thus, we have a map from $B$ to the pullback of $i\_0$ and $i\_1$. Since coproducts in an extensive category are disjoint, this pullback is an initial object. Since initial objects in an extensive category are strict, $B$ is also initial.


