Proposition 10.9.10. Let $\overline{S}$ be the image of $S$ in $S'^{-1}A$, then $(SS')^{-1}A$ is isomorphic to $\overline{S}^{-1}(S'^{-1}A)$.

Proof. The map sending $x\in A$ to $x/1\in (SS')^{-1}A$ induces a map sending $x/s\in S'^{-1}A$ to $x/s \in (SS')^{-1}A$, by universal property. The image of the elements in $\overline{S}$ are invertible in $(SS')^{-1}A$. By the universal property we get a map $f : \overline{S}^{-1}(S'^{-1}A) \to (SS')^{-1}A$ which maps $(x/t')/(s/s')$ to $(x/t')\cdot (s/s')^{-1}$.

On the other hand, the map from $A$ to $\overline{S}^{-1}(S'^{-1}A)$ sending $x\in A$ to $(x/1)/(1/1)$ also induces a map $g : (SS')^{-1}A \to \overline{S}^{-1}(S'^{-1}A)$ which sends $x/ss'$ to $(x/s')/(s/1)$, by the universal property again. It is immediately checked that $f$ and $g$ are inverse to each other, hence they are both isomorphisms. $\square$

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