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?


1 Answer 1


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.

  • $\begingroup$ Thank you for the presented argument. I carelessly imagined defining the "set of binary coproduct decompositions" as the set of equivalence classes of such things, but perhaps that's ill-advised. The assignment I imagined to be bijective was between the third diagram(s) of Proposition 4.1 in Carboni and Lack's paper and its middle arrow(s). $\endgroup$
    – Arrow
    Dec 11, 2017 at 22:12
  • $\begingroup$ @Arrow I don't think that's ill-advised at all. In an extensive category, coproduct decompositions form a discrete groupoid, so it's equivalent to a set. $\endgroup$ Dec 12, 2017 at 3:01

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