Proposition 10.9.11. View $S'^{-1}M$ as an $A$-module, then $S^{-1}(S'^{-1}M)$ is isomorphic to $(SS')^{-1}M$.

**Proof.**
Note that given a $A$-module M, we have not proved any universal property for $S^{-1}M$. Hence we cannot reason as in the preceding proof; we have to construct the isomorphism explicitly.

We define the maps as follows

We have to check that these homomorphisms are well-defined, that is, independent the choice of the fraction. This is easily checked and it is also straightforward to show that they are inverse to each other. $\square$

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