When is a scheme, considered as a functor $X: \underline{k-Alg} \to \underline{Set}$, connected?

Question

Let $k$ be a commutative ring and $X$ a $k$-scheme. Then $X$ is connected iff $X$ is non-empty and for all open subsets $Y_1,Y_2$ of $X$: $X=Y_1 \sqcup Y_2 \Rightarrow Y_1=X$ or $Y_2=X$.

Now we can also consider $X$ as a functor $\bar X :\underline{k-Alg} \to \underline{Set}$ and open subschemes $Y\subset X $ correspond to open subfunctor $\bar Y\subset \bar X$.

How does the connectedness condition of the scheme $X$ translate to the view as a k-functor? Are there any conditions easily verifiable on $\bar X :\underline{k-Alg} \to \underline{Set}$, which induce connectedness of $X$?

Answer 1

This is a trivial reformulation, but ... $X$ is connected iff the functor $h_X : \mathbf{CAlg}_k \to \mathbf{Set}$ has the following properties:

1. $h_X$ is not the functor $h_{\emptyset}$, which is $h_{\emptyset}(0)=\star$ and $h_{\emptyset}(R)=\emptyset$ for $R \neq 0$
2. If $h_X = F \vee G$ for two disjoint open subfunctors $F,G$, then $h_X = F$ or $ h_X = G$.

Here, a subfunctor $F \to F'$ is called open when for every $R$-point $s \in F'(R)$ the pullback $F \times_{F'} h_R \to h_R$ is represented by an open subscheme of $\mathrm{Spec}(R)$, and $F \vee F'$ is the smallest open subfunctor containing $F$ and $F'$, and $F,F'$ are disjoint when $F(K) \cap F'(K)=\emptyset$ for all fields $K$.