For an extensive category, the following conditions are equivalent for an object $C$.
- The representable copresheaf of $C$ commutes with coproducts.
- The $C=X\amalg Y\implies X\text{ or }Y$ is $0$ and $C\neq 0$
How to prove these are equivalent?
Following Zhen Lin's hint for the $\Uparrow$ direction, I tried looking at the diagram below. One of the characterizations of extensivity says the top row must now be a coproduct. By connectedness, one of the top coproduct injections must be an iso. Suppose wlog the left one is an iso (as in the diagram). Then $\mathrm{Hom}(A\times_{A\amalg B}C,A+B)\cong \mathrm{Hom}(C,A+B)$. However, I want to LHS to be $\mathrm{Hom}(A\times_{A\amalg B}C,A)$, and I'm not sure how to get there.
$$\require{AMScd} \begin{CD} A\times_{A\amalg B}C @>{\cong}>> C @<<< B\times_{A\amalg B}C\\ @VVV @V{f}VV @VVV \\ A @>>> A+B @<<< B \end{CD}$$
How to finish this direction?
This is the proof of theorem 2 here. For $\Uparrow$ use the diagram below noting that coproduct injections are monos in extensive categories. The proof is spelled out in the link.
$$\require{AMScd} \begin{CD} A\times_{A\amalg B}C @>>> C @<<< B\times_{A\amalg B}C\\ @VVV @V{f}VV @VVV \\ A @>>> A+B @<<< B \end{CD}$$
The $\Downarrow$ direction is basically "$1_C$ can't have a preimage".