Existence of Homomorphism between Localizations of Ring

Theorem
Let $A$ be a commutative ring with unity.

Let $S, T \subseteq A$ be multiplicatively closed subsets.


 * $(1): \quad$ There exists an $A$-algebra homomorphism $h : A_S \to A_T$ between localizations, the induced homomorphism.
 * $(2): \quad S$ is a subset of the saturation of $T$.
 * $(3): \quad$ The saturation of $S$ is a subset of the saturation of $T$.
 * $(4): \quad$ Every prime ideal meeting $S$ also meets $T$.

Proof
Let $i : A \to A_S$ and $j : A \to A_T$ be the localization homomorphisms.

1 implies 2
Let $h : A_S \to A_T$ be an $A$-algebra homomorphism.

Then by definition, $j = h \circ i$:
 * $\xymatrix{

A \ar[d]_i \ar[r]^{j} & A_T\\ A_S \ar[ru]_{h} }$

Let $s \in S$.

By definition of localization, $i(s)$ is a unit of $A_S$.

By Ring Homomorphism Preserves Invertible Elements, $j(s) = h(i(s))$ is a unit of $A_T$.

Thus $s$ is an element of the saturation of $T$.

2 implies 1
Let $S$ be a subset of the saturation of $T$.

Then its image $j(S) \subseteq A_T^\times$ consists of units of $A_T$.

By definition of localization at $S$, there exists a unique $A$-algebra homomorphism $h : A_S \to A_T$.

2 implies 3
Let $S$ be a subset of the saturation of $T$.

By definition, its saturation is the smallest saturated multiplicatively closed subset of $A$ containing $S$.

Thus the saturation of $S$ is a subset of the saturation of $T$.

3 implies 2
By definition, $S$ is a subset of its saturation.

Also see

 * Uniqueness of Homomorphism between Localizations of Ring