We assume all rings are commutative. Let $A$ be a ring. Let $I$ be a small filtered category. Let $F\colon I \rightarrow A$-alg be a functor, where $A$-alg is the category of $A$-algebras. We write $X = Spec(A)$, $B_i = F(i)$ for each $i \in I$, $B =$ colim $B_i$, $Y_i = Spec(B_i)$ for each $i\in I$, $Y = Spec(B)$. Let $f\colon Y \rightarrow X$, $f_i\colon Y_i \rightarrow X$ be the canonical morphisms.
Then $f(Y) = \bigcap f_i(Y_i)$?
