# Localization of the pullback diagram

In the paper, Topologically Defined Classes of Commutative Rings, localization of the pullback diagram (with $v,$ surjective) $$\begin{array} DD & \stackrel{v\ '}{\longrightarrow} & A \\ \downarrow{u' } & & \downarrow{u} \\ B & \stackrel{v}{\longrightarrow} & C \end{array}$$ is given as Proposition 1.9:
... Conversely, if $S_A$ is a multiplicatively closed set of $A$ and if $S_B$ is a multiplicatively closed of $B$ and if $u(S_A)=v(S_B)=S_C$ then $S_A^{-1}A \times_{S_C^{-1}C}S_B^{-1}B \cong (S_A \times_{S_C}S_B)^{-1} D.$

I can not see how a typical element of one side mapped to the other side by this isomorphism.
First, you expect that the pullback (as a special kind of limit) should be the target of your map. Thus you expect a map: $$(S_A \times_{S_C}S_B)^{-1} D\longrightarrow S_A^{-1}A \times_{S_C^{-1}C}S_B^{-1}B$$ Denote $S=S_A \times_{S_C}S_B$. In this setting you have $S=v'^{-1}(S_A)\cap u'^{-1}(S_B)$. So that it is naturally a multiplicative subset of $D$. In your context the localization $S^{-1}D$ is the set of pairs or formal quotients $d/s$ modulo identifications. Then $S_A^{-1}A \times_{S_C^{-1}C}S_B^{-1}B\subseteq S_A^{-1}A \times S_B^{-1}B$ is a subset so that you define: $$d/s\mapsto (v'(d)/v'(s),u'(d)/u'(s))$$ and you verify, that this agrees with the appropriate identifications and that both elements of the pair on the right are mapped to the same element of $S_C^{-1}C$.