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/s')/(s/1) to x/ss'.
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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