Proposition 10.9.3. Let $f : A \to B$ be a ring map that sends every element in $S$ to a unit of $B$. Then there is a unique homomorphism $g : S^{-1}A \to B$ such that the following diagram commutes.
Proof. Existence. We define a map $g$ as follows. For $x/s\in S^{-1}A$, let $g(x/s) = f(x)f(s)^{-1}\in B$. It is easily checked from the definition that this is a well-defined ring map. And it is also clear that this makes the diagram commutative.
Uniqueness. We now show that if $g' : S^{-1}A \to B$ satisfies $g'(x/1) = f(x)$, then $g = g'$. Hence $f(s) = g'(s/1)$ for $s \in S$ by the commutativity of the diagram. But then $g'(1/s)f(s) = 1$ in $B$, which implies that $g'(1/s) = f(s)^{-1}$ and hence $g'(x/s) = g'(x/1)g'(1/s) = f(x)f(s)^{-1} = g(x/s)$. $\square$
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