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

- In a coproduct decomposition, exactly one of the summands is initial;
- 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?