Let $ A $ be a ring.
Let $ I $ be a preordered set, filtering.
Let $ \Sigma $ a multiplicative subset of $ A $.
Suppose for any given $ i \in I $ a multiplicative subset $ S_i $ of $ A $ contained in $ \Sigma $.
Make the following assumptions:
$ \Sigma = \displaystyle \bigcup_i S_i $
For $ (i, j) \in I^2 $ with $ i \leq j $, elements $ S_i $ become invertible in $ S_ {j}^{-1} A $.
$ \mathcal{D} = ((S_{i}^{-1} A)_{i \in I} \ , \ (S_{i}^{-1} A \to S_{j}^{-1} A)_{i \leq j}) $ is a commutative diagram in the category of filter $ A $ - algebras.
Question :
Show that $ \displaystyle \lim_{\longrightarrow} \mathcal{D} \simeq \Sigma^{-1} A $.
Thank you in advance for your help.
$\Sigma^{-1} A$ represents the functor of homomorphisms on $A$ which invert all elements in $\Sigma$. Since $\Sigma$ is covered by the $S_i$, this means that all the $S_i$ are made invertible. Hence, we get unique lifts to $S_i^{-1} A$, and these are clearly compatible. The construction is invertible, so that $\Sigma^{-1} A$ represents the same functor as $\varinjlim_i S_i^{-1}A$, i.e. they are isomorphic (Yoneda Lemma).