I want to show the following. Let $X$ be any scheme (say over the terminal object $\operatorname{Spec} \Bbb{Z}$ in $\textbf{Sch}$) and $A \to B$ a faithfully flat ring homomorphism. Then
$$h_X(A) \to h_X(B) \stackrel{\longrightarrow}{_\longrightarrow} h_X( B\otimes_A B)$$
is an equalizer. I am being lazy here and denoting $\operatorname{Spec} A$ by just $A$, etc. Now the case when $X$ is affine is easy; this follows from the fact that the relevant Amitsur complex is acyclic when $A \to B$ is faithfully flat. However how can we reduce to the case that $X$ is affine? I have tried many things like assume $X$ is separated and $A \to B$ finite type to see if this works, and can come up with something. Is this result though true in general?
The claim is true, but I don't know of any easy way of reducing to the affine case. There is some real work to be done here, in my opinion! See e.g. Theorem 2.55 in Vistoli's notes.