In the paper, [Topologically Defined Classes of Commutative Rings](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=0ahUKEwj47P6Zp9_RAhXFFiwKHabcAB8QFgghMAE&url=http%3A%2F%2Fwww.mat.uniroma3.it%2Fusers%2Ffontana%2Fdocuments%2FFontana-AnnMatPA-80.pdf&usg=AFQjCNGIKDglVjFV737cTRGhxjXD1UDq8A&sig2=dbJsR7tr9dJ25EBAb5HwUw), 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.  

Can you help please?  

Thank you.



Localization of the pullback diagram
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 $S\_C^{-1}C$.


