Let $f\colon X\to Y$ be a morphism of schemes.
Let $\operatorname{Spec}A,\operatorname{Spec}C$ be affine open subschemes of $Y$ such that $\operatorname{Spec}A_g=\operatorname{Spec}C_f$ for some $g \in A$,$f\in C$. Let $\operatorname{Spec}B, \operatorname{Spec}D$ be affine open subschemes of $X$ such that $f^{-1}(\operatorname{Spec}A)=\operatorname{Spec}B$ and $f^{-1}(\operatorname{Spec}C)=\operatorname{Spec}D$. This implies $\operatorname{Spec}B_{f^\sharp_{\operatorname{Spec}A}(g)}= \operatorname{Spec}D_{f^\sharp_{\operatorname{Spec}C}(f)}$. Thus, there exist isomorphisms $A_g \simeq C_f$ and $B_{f^\sharp_{\operatorname{Spec}A}(g)}\simeq D_{f^\sharp_{\operatorname{Spec}C}(f)}$. I think it's true that these isomorphisms make the diagram
$$
\require{AMScd}
\begin{CD}
A_g @>{f^\sharp_{\operatorname{Spec}A_g}}>> B_{f^\sharp_{\operatorname{Spec}A}(g)}\\
@VVV @VVV \\
C_f @>{f^\sharp_{\operatorname{Spec}C_f}}>> D_{f^\sharp_{\operatorname{Spec}C}(f)}
\end{CD}
$$
commute but I can't come up with a proof.
I am grateful for any kind of information about this.